I found this nice explicit formula for the Bernoulli numbers:
$$B_n = \sum_{k \mathop = 0}^n \sum_{i \mathop = 0}^k (-1)^i \binom k i \frac {i^n} {k + 1}$$
I can't find a proof though. I want to prove it from the generating function definition:
$$ \frac x {e^x - 1} = \sum_{n \mathop = 0}^\infty B_n \frac {x^n} {n!}$$
Any proof sketches or links will be appreciated.
So as not to get mixed up with complex variables use $j$ rather than $i$ to get $$\sum_{k=0}^n \sum_{j=0}^k (-1)^j {k\choose j} \frac{j^n}{k+1}.$$ This is $$\sum_{k=0}^n \frac{1}{k+1} \sum_{j=0}^k (-1)^j {k\choose j} j^n = \sum_{k=0}^n \frac{1}{k+1} (-1)^k \times k! \times {n\brace k}.$$
Recall the classic generating function of the Stirling numbers of the second kind which yields $${n\brace k} = n! [z^n][u^k] \exp(u(\exp(z)-1)).$$ Substituting this into the sum gives $$n![z^n] \sum_{k=0}^n \frac{1}{k+1} (-1)^k \times k! \times [u^k] \exp(u(\exp(z)-1)) \\ = n![z^n] \sum_{k=0}^n \frac{1}{k+1} (-1)^k \times k! \times \frac{(\exp(z)-1)^k}{k!} \\= n![z^n] \sum_{k=0}^n \frac{1}{k+1} (-1)^k \times (\exp(z)-1)^k$$ Now observe that $\exp(z)-1$ starts at $z$ and hence we can extend the summation to infinity without affecting $[z^n]$ to get $$n![z^n] \sum_{k=0}^\infty \frac{1}{k+1} (-1)^k \times (\exp(z)-1)^k \\=n! [z^n] \frac{1}{\exp(z)-1} \sum_{k=0}^\infty \frac{1}{k+1} (-1)^k \times (\exp(z)-1)^{k+1} \\= n! [z^n] \frac{1}{\exp(z)-1} \log(1+\exp(z)-1) = n! [z^n] \frac{z}{\exp(z)-1}.$$ Done.
Nice how Bernoulli numbers show up in both analytic number theory and combinatorics.
There have been many references collected for proofs of this formula of the Bernoulli numbers $B_n$ at https://math.stackexchange.com/a/4254493/945479.
The explicit formula \begin{equation}\label{Higgins-Gould-B}\tag{1} B_n=\sum_{k=0}^n\frac1{k+1}\sum_{j=0}^k(-1)^j\binom{k}{j}j^n,\quad n\ge0, \end{equation} has a long history, it appeared in the paper
It is a special case of the formula (2.5) in the paper
Its equivalent form is \begin{equation}\label{Bernoulli-Stirling-eq}\tag{2} B_n=\sum_{k=0}^n(-1)^k\frac{k!}{k+1}S(n,k), \quad n\ge0, \end{equation} where $S(n,k)$ is the Stirling numbers of the second kind. See page 29, Remark 2 in the paper
There existed at least seven alternative proofs of the formulas \eqref{Higgins-Gould-B} and \eqref{Bernoulli-Stirling-eq} in the paper [1, 2] above and in the following monographs and papers:
To the best of my knowledge, except the formulas \eqref{Higgins-Gould-B} and \eqref{Bernoulli-Stirling-eq} above, there are also the following explicit formulas for the Bernoulli numbers $B_n$: \begin{align}\label{Higgins-Gould-B(11)}\tag{3} B_n&=\sum_{j=0}^n(-1)^j\binom{n+1}{j+1}\frac{n!}{(n+j)!}\sum_{k=0}^j(-1)^{j-k}\binom{j}{k}k^{n+j}, \quad n\ge0;\\ B_n&=\sum_{i=0}^n(-1)^{i}\frac{\binom{n+1}{i+1}}{\binom{n+i}{i}}S(n+i,i), \quad n\ge0;\label{Bernoulli-Stirling-formula}\tag{4}\\ B_{2k}&=1+\sum_{m=1}^{2k-1}\frac{S(2k+1,m+1) S(2k,2k-m)}{\binom{2k}{m}}\\ &\quad-\frac{2k}{2k+1}\sum_{m=1}^{2k}\frac{S(2k,m)S(2k+1,2k-m+1)}{\binom{2k}{m-1}}, \quad k\in\mathbb{N};\tag{5}\\ B_{2k}&=\frac{(-1)^{k-1}k}{2^{2(k-1)}(2^{2k}-1)}\sum_{i=0}^{k-1}\sum_{\ell=0}^{k-i-1} (-1)^{i+\ell}\binom{2k}{\ell}(k-i-\ell)^{2k-1}, \quad k\in\mathbb{N};\tag{6}\\ B_{2m}&=(-1)^{m-1}\frac{m} {2^{2m-1}\bigl(2^{2m}-1\bigr)}\Biggl[\sum_{k=0}^{m-1} (-1)^k\binom{2m}{k}(m-k)^{2m-1}\\ &\quad+2\sum_{k=1}^{m-1}(-1)^k\sum_{\ell=0}^{m-k-1} (-1)^{\ell}\binom{2m}{\ell}(m-k-\ell)^{2m-1}\Biggr],\quad m\in\mathbb{N};\\ B_{2m}&=\frac{m} {2^{2m-1}\bigl(2^{2m}-1\bigr)}\sum_{\ell=1}^{2m}\frac{(-1)^{\ell-1}}{2^\ell} \biggl(\frac1\ell-\frac1{m+1}\biggr) \binom{2m+1}{\ell} \sum_{q=0}^\ell\binom{\ell}{q}(2q-\ell)^{2m}, \quad m\in\mathbb{N};\\ B_{2k}&= \frac12 - \frac1{2k+1} - 2k \sum_{i=1}^{k-1} \frac{A_{2(k-i)}}{2(k - i) + 1},\quad k\in\mathbb{N};\tag{7} \end{align} where $A_m$ is defined by \begin{equation*} \sum_{m=1}^nm^k=\sum_{m=0}^{k+1}A_mn^{m}. \end{equation*} The formulas \eqref{Higgins-Gould-B(11)} and \eqref{Bernoulli-Stirling-formula} are also equivalent to eah other.
By the way, I would like to mention two intereting double inequalities related to the Bernoulli numbers $B_{2n}$ as follows.
More related references
Let $\mathbb{N}=\{1,2,\dotsc,\}$ be the set of the positive integers.
In the paper [2] below, the Bernoulli polynomials $B_n(x)$ were determinantally expressed as \begin{equation} B_n(x)=\frac{(-1)^n}{(n-1)!} \begin{vmatrix} 1 & x & x^2 & x^3 & \dotsm & x^{n-1} & x^n\\ 1 & \frac12 & \frac13 & \frac14 & \dotsm & \frac1n & \frac1{n+1}\\ 0 & 1 & 1 & 1 & \dotsm & 1 & 1\\ 0 & 0 & 2 & 3 & \dotsm & n-1 & n\\ 0 & 0 & 0 & \binom32 & \dotsm & \binom{n-1}2 & \binom{n}2\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & \dotsm & \binom{n-1}{n-2} & \binom{n}{n-2} \end{vmatrix}, \quad n\in\mathbb{N}. \end{equation} In the paper [1] below, the Bernoulli polynomials $B_n(x)$ were represented as \begin{align*} B_n(x)&=(-1)^nn! \begin{vmatrix} 1 & 1 & 0 & 0 & 0 & 0 & \dotsm & 0\\ \frac{x}{1!} & \frac1{2!} & 1 & 0 & 0 & 0 & \dotsm & 0\\ \frac{x^2}{2!} & \frac1{3!} & \frac1{2!} & 1 & 0 & 0 & \dotsm & 0\\ \frac{x^3}{3!} & \frac1{4!} & \frac1{3!} & \frac1{2!} & 1 & 0 & \dotsm & 0\\ \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm\\ \frac{x^n}{n!} & \frac1{(n+1)!} & \frac1{n!} & \frac1{(n-1)!} & \frac1{(n-2)!}&\frac1{(n-3)!}&\dotsm& 1 \end{vmatrix}\\ &=(-1)^n\frac{n!}{\prod_{k=1}^nk!} \begin{vmatrix} 1 & 1 & 0 & 0 & 0 & 0 & \dotsm & 0\\ x & \frac1{2!} & 1 & 0 & 0 & 0 & \dotsm & 0\\ x^2 & \frac{2!}{3!} & 1 & 2! & 0 & 0 & \dotsm & 0\\ x^3 & \frac{3!}{4!} & 1 & \frac{3!}{2!} & 3! & 0 & \dotsm & 0\\ \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm\\ x^n & \frac{n!}{(n+1)!} & 1 & \frac{n!}{(n-1)!} & \frac{n!}{(n-2)!}&\frac{n!}{(n-3)!}&\dotsm& n! \end{vmatrix}\\ &=(-1)^n\prod_{k=1}^{n-1}\frac{(k-1)!}{k!} \begin{vmatrix} 1 & 1 & 0 & 0 & 0 & 0 & \dotsm & 0\\ x & \frac1{2!} & 1 & 0 & 0 & 0 & \dotsm & 0\\ x^2 & \frac{2!}{3!} & 1 & 2! & 0 & 0 & \dotsm & 0\\ x^3 & \frac{3!}{4!} & 1 & \frac{3!}{2!} & \frac{3!}{2!} & 0 & \dotsm & 0\\ \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm\\ x^n & \frac{n!}{(n+1)!} & 1 & \frac{n!}{(n-1)!} & \frac{n!}{(n-2)!2!}&\frac{n!}{(n-3)!3!}&\dotsm& \frac{n!}{2!(n-2)!} \end{vmatrix}. \end{align*} In the paper [4], the Bernoulli polynomials $B_n(x)$ was explicitly expressed as \begin{multline}\label{BP-Stirl-form} B_n(x)=\sum_{k=1}^nk!\sum_{r+s=k}\sum_{\ell+m=n}(-1)^m\binom{n}{\ell} \frac{\ell!}{(\ell+r)!}\frac{m!}{(m+s)!}\\ \times\Biggl[\sum_{i=0}^r\sum_{j=0}^s(-1)^{i+j}\binom{\ell+r}{r-i}\binom{m+s}{s-j}S(\ell+i,i)S(m+j,j)\Biggr]x^{m+s}(1-x)^{\ell+r} \end{multline} and was determinantally represented as \begin{equation}\label{Bern-Polyn-determ} B_n(x)=(-1)^n\biggl|\frac1{\ell+1}\binom{\ell+1}{m} \bigl[(1-x)^{\ell-m+1}-(-x)^{\ell-m+1}\bigr]\biggr|_{1\le \ell\le n,0\le m\le n-1} \end{equation} for $n\in\mathbb{N}$, where $S(n,k)$ denotes the Stirling numbers of the second kind and $|\cdot|_{1\le \ell\le n,0\le m\le n-1}$ denotes a $n\times n$ determinant.
In the paper [5], an alternative determinantal expression \begin{equation}\label{Bernoulli-Polyn-Det-Erew} B_n(x)=\frac{(-1)^n}{(n+1)!} \begin{vmatrix} 1&1&0&0&\dotsm&0&0&0\\ \frac{x}2&\binom{2}0&\frac12&0&\dotsm&0&0&0\\ \frac{x^2}3&\binom{3}0&\binom{3}1&\frac13&\dotsm&0&0&0\\ \frac{x^3}4&\binom{4}0&\binom{4}1&\binom{4}2&\dotsm&0&0&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ \frac{x^{n-2}}{n-1}&\binom{n-1}0&\binom{n-1}1&\binom{n-1}2&\dotsm &\binom{n-1}{n-3}&\frac1{n-1}&0\\ \frac{x^{n-1}}{n}&\binom{n}0&\binom{n}1&\binom{n}2&\dotsm &\binom{n}{n-3}&\binom{n}{n-2}&\frac1{n}\\ \frac{x^n}{n+1}&\binom{n+1}0&\binom{n+1}1&\binom{n+1}2&\dotsm &\binom{n+1}{n-3}&\binom{n+1}{n-2}&\binom{n+1}{n-1} \end{vmatrix}, \quad n\ge0 \end{equation} for the Bernoulli numbers $B_n(x)$ was derived.
Theorem 1.1 in [3] reads that, for all integers $n,r\ge0$, the Bernoulli polynomials $B_n(r)$ can be computed in terms of the $r$-Stirling numbers of the second kind $S_r(n,k)$ by \begin{equation}\label{Bernoulli-Poly-r-Stirling-eq} B_n(r)=\sum_{k=0}^n(-1)^k\frac{k!}{k+1}S_r(n,k). \end{equation}
When taking $x=0$ in the above determinatal expressions and closed-form expressions, we can arrive at closed-form formulas for the Bernoulli numbers $B_n=B_n(0)$ in terms of the determinants of the Hessenberg matrices.
References
Starting with the definition of Bernouilli number by Bernouilli polynomials $$B_n=\left.\frac D{e^D-1}x^n\right\vert_{x=0}$$ $$B_n=\left.\frac{\ln(1+\Delta)}{\Delta}x^n\right\vert_{x=0}$$ where $\Delta$ is the forward difference operator, $$B_n=\left.\sum_{k=0}^\infty\frac{(-1)^k}{k+1}\Delta^kx^n\right\vert_{x=0}$$ $$B_n=\left.\sum_{k=0}^\infty\frac{(-1)^k}{k+1}\Delta^kx^n\right\vert_{x=0}$$ $$B_n=\left.\sum_{k=0}^n\frac{(-1)^k}{k+1}\sum_{i=0}^k(-1)^{k-i}{k\choose i}(x+i)^n\right\vert_{x=0}$$ we reach the expression in the title.
This is a very elementary proof of Bernoulli numbers formula (for more details see https://www.pulsus.com/scholarly-articles/a-very-elementary-proof-of-an-explicit-formula-of-bernoulli-numbers.pdf):
The explicit formula of Stirling numbers of the second kind is given by:
\begin{equation} k!S(n,k)=\sum_{i=0}^{k}(-1)^{k-i}\binom{k}{i}i^{n} \end{equation}
If we invert the above formula using the binomial inversion formula (see https://en.wikipedia.org/wiki/Binomial_transform), we get:
$$k^{n}=\sum_{i=0}^{k}S(n,i){(k)}_{i}$$ valid for $k\in \{0, 1, \cdots, n\}$ and $(x)_n=x(x-1)\cdots(x-n+1)$.
Now sum for $k$ in the last formula you get:
$$ \begin{split} \sum_{k=0}^{n}k^{n} & = \sum_{k=0}^{n}\sum_{i=0}^{k}S(n,i){(k)}_{i} \\ &=\sum_{i=0}^{n}S(n,i)\sum_{k=i}^{n}{(k)}_{i} \\ &=\sum_{i=0}^{n}S(n,i)\left(\frac{{(n+1)}_{i+1}-{(i)}_{i+1}}{i+1}\right) \\ &=\sum_{i=0}^{n}S(n,i)\frac{{(n+1)}_{i+1}}{i+1}\\ &=\sum_{i=0}^{n}S(n,i)\frac{1}{i+1}\sum_{j=0}^{i}c_{j}(n+1)^{j+1} \\ &=\sum_{j=0}^{n}\left(\sum_{i=j}^{n}S(n,i)\frac{c_{j}}{i+1}\right)(n+1)^{j+1} \end{split} $$
Thus: $$\sum_{k=0}^{n}k^{n}=\sum_{j=0}^{n}\left(\sum_{i=j}^{n}S(n,i)\frac{c_{j}}{i+1}\right)(n+1)^{j+1}$$
Using Faulhaber's formula for the LHS of the above formula you get:
$$\sum_{j=0}^{n}\left(\frac{\binom{n+1}{j+1}}{n+1}B_{n-j}\right)(n+1)^{j+1}=\sum_{j=0}^{n}\left(\sum_{i=j}^{n}S(n,i)\frac{c_{j}}{i+1}\right)(n+1)^{j+1}$$
The coefficients of $(n+1)$ in both representations are equal so: $$\frac{\binom{n+1}{1}}{n+1}B_{n}=\sum_{i=0}^{n}S(n,i)\frac{c_{0}}{i+1} \implies B_{n}=\sum_{i=0}^{n}S(n,i)\frac{{(-1)}^{i}i!}{i+1}$$
Substitute $i!S(n,i)$ by the very first formula you get the famous explicit formula of Bernoulli numbers:
$$B_{n}=\sum_{i=0}^{n}\frac{1}{i+1}\sum_{j=0}^{i}\binom{i}{j}{(-1)}^{j}j^{n}$$
