We seek to show that with $0\le k\le n$ the following identity holds:
$$\sum_{j=0}^k (-1)^{k-j} {2n+1\choose k-j} {n+j\brace j} =
\sum_{j=0}^{n-k} (-1)^j {2n+1\choose j} {2n-k-j+1\brack n-k-j+1}.$$
We will start with the LHS. The chapter 6.2 on Eulerian Numbers
of Concrete Mathematics by Knuth et al. proposes the formula
$${n\brace m} = (-1)^{n-m+1} \frac{n!}{(m-1)!} \sigma_{n-m}(-m)$$
where $\sigma_n(x)$ is a Stirling polynomial and we have the identity
$$\left(\frac{1}{z} \log\frac{1}{1-z}\right)^x
= x \sum_{n\ge 0} \sigma_n(x+n) z^n.$$
We get
$$[z^{n-m}] \left(\frac{1}{z} \log\frac{1}{1-z}\right)^x
= x \sigma_{n-m}(x+n-m)$$
and hence
$$[z^{n-m}] \left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n}
= -n \sigma_{n-m}(-m)$$
which implies that for $n\ge m\ge 1$
$$\bbox[5px,border:2px solid #00A000]{
{n\brace m} = (-1)^{n-m} \frac{(n-1)!}{(m-1)!}
[z^{n-m}] \left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n}.}$$
This gives for the LHS
$$\sum_{j=1}^k (-1)^{k-j} {2n+1\choose k-j}
(-1)^n \frac{(n+j-1)!}{(j-1)!} [z^n]
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n-j}
\\ = (-1)^{n-k+1} n! [z^n]
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n-1}
[w^{k-1}] (1+w)^{2n+1}
\\ \times \sum_{j=1}^k {n+j-1\choose n} (-1)^{j-1} w^{j-1}
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-j+1}.$$
Now the coefficient extractor in $w$ enforces the upper limit of the
sum and we may extend $j$ to infinity, getting
$$(-1)^{n-k+1} n! [z^n]
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n-1}
[w^{k-1}] (1+w)^{2n+1}
\frac{1}{(1+w/(\frac{1}{z} \log\frac{1}{1-z}))^{n+1}}
\\ = (-1)^{n-k+1} n! [z^n]
[w^{k-1}] (1+w)^{2n+1}
\frac{1}{(w+\frac{1}{z} \log\frac{1}{1-z})^{n+1}}.$$
Continuing,
$$ (-1)^{n-k+1} n! [z^n]
[w^{n+k}] (1+w)^{2n+1}
\frac{1}{(1+\frac{1}{w} \frac{1}{z} \log\frac{1}{1-z})^{n+1}}
\\ = (-1)^{n-k+1} n! [z^n]
[w^{n+k}] (1+w)^{2n+1}
\sum_{q\ge 0} {n+q\choose n} (-1)^q \frac{1}{w^q}
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^q
\\ = (-1)^{n-k+1} n! [z^n]
\sum_{j=n+k}^{2n+1} {2n+1\choose j}
{n+j-(n+k)\choose n} (-1)^{j-(n+k)}
\\ \times
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{j-(n+k)}
\\ = (-1)^{n-k+1} n! [z^n]
\sum_{j=0}^{n-k+1} {2n+1\choose j+n+k}
{n+j\choose n} (-1)^{j}
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{j}
\\ = (-1)^{n-k+1} n!
\sum_{j=0}^{n-k+1} {2n+1\choose j+n+k}
{n+j\choose n} (-1)^{j} [z^{n+j}]
\left(\log\frac{1}{1-z}\right)^{j}
\\ = (-1)^{n-k+1} n!
\sum_{j=0}^{n-k+1} {2n+1\choose j+n+k}
{n+j\choose n} (-1)^{j}
\\ \times \frac{j!}{(n+j)!} \times (n+j)! [z^{n+j}]
\frac{1}{j!} \left(\log\frac{1}{1-z}\right)^{j}
\\ = (-1)^{n-k+1} \sum_{j=0}^{n-k+1}
{2n+1\choose j+n+k} (-1)^j {n+j\brack j}
\\ = (-1)^{n-k+1} \sum_{j=0}^{n-k+1}
{2n+1\choose 2n-j+1} (-1)^{n-k-j+1} {2n-k-j+1\brack n-k-j+1}
\\ = \sum_{j=0}^{n-k+1}
{2n+1\choose j} (-1)^j {2n-k-j+1\brack n-k-j+1}.$$
The Stirling number is zero for $j=n-k+1$ and we get at last
$$\sum_{j=0}^{n-k}
{2n+1\choose j} (-1)^j {2n-k-j+1\brack n-k-j+1}.$$
This is the RHS and we have the claim.
