In the derivation of the Bernoulli-Seki formula the first step is unfolding the exponential generating function of $S_n(k)=\sum_{a=1}^k a^n$ as $$ \sum_{n \geq 0} S_n(k) \frac{T^n}{n !}=\sum_{a=1}^k \sum_{n \geq 0} \frac{(a T)^n}{n !}=\sum_{a=1}^k e^{a T}=e^T \frac{e^{k T}-1}{e^T-1} $$
This part is clear.
However, I don't follow the equality
$$\frac{T e^T}{e^T-1}=\sum_{m \geq 0} B_m \frac{T^m} {m !}$$
in
$$ e^T \frac{e^{k T}-1}{e^T-1}=\frac{T e^T}{e^T-1} \cdot \frac{e^{k T}-1}{T}=\left(\sum_{m \geq 0} B_m \frac{T^m}{m !}\right)\left(\sum_{n \geq 0} \frac{k^{n+1} T^n}{(n+1) !}\right) $$
with $B_m$ being the corresponding Bernoulli number.
I get that it is a definition, but the coefficients (Bernoulli numbers) are already embedded in an algebraic expression in the RHS. How is it guaranteed that such a generating function exists at all?
I see now that
$$\frac{e^{kT}-1}{T}=\sum_{n \geq 0} \frac{k^{n+1} T^n}{(n+1)!}\\=\sum_{n \geq 0} \frac{k^{n+1} T^{n+1}}{(n+1)!}\frac 1 T=\left(e^{kT}-1\right)\frac 1 T$$
The question is answered here.
