Bernoulli numbers and the sum of the $m$-th powers of the first $n$ integers 
Let $S_m(x)$ denote the following polynomial in $x$
$$S_m(x) = \sum_{k=0}^m \frac1{k+1}\cdot\binom mk\cdot B_{m-k}\cdot x^{k+1}$$
Prove that $$S_m(x+1)-S_m(x) = x^m$$ for all $m >0$. Use this to show that $$(-1)^m+(-2)^m+\cdots+(-n)^m = -S_m(-n)$$ for all $n >0$ and $m>0$.

I'm really stuck on this question. I don't know how to prove the first part or use it to show the second part. Any help would be much appreciated.
 A: The first part can be done using generating functions, quite a nice exercise in manipulation thereof. Rewrite the sum as follows:
$$S_m(x) = \sum_{k=0}^m {m\choose k} \frac{x^{k+1}}{k+1} B_{m-k}$$
where the $B_q$ are the Bernoulli  numbers.
Observe that when we multiply two exponential generating functions of the sequences $\{a_n\}$ and $\{b_n\}$ we get that
$$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!} \sum_{n\ge 0} b_n \frac{z^n}{n!}
= \sum_{n\ge 0} \sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\
= \sum_{n\ge 0} \sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!}
= \sum_{n\ge 0} \left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$
i.e. the product of the two generating functions is the generating function of
$$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$
Now we clearly have
$$A(z) = \sum_{n\ge 0} \frac{x^{n+1}}{n+1} \frac{z^n}{n!}
= \frac{1}{z}  \sum_{n\ge 0} x^{n+1} \frac{z^{n+1}}{(n+1)!}
= \frac{1}{z} (\exp(xz)-1).$$
The exponential generating function of the Bernoulli numbers is
$$B(z) = \frac{z}{\exp(z)-1}.$$
Therefore the sum is given by
$$m![z^m] A(z) B(z) = m! [z^m]\frac{1}{z} (\exp(xz)-1) \frac{z}{\exp(z)-1}
=  m! [z^m] \frac{\exp(xz)-1}{\exp(z)-1}.$$
The difference we are trying to evaluate is thus
$$m! [z^m] \frac{\exp((x+1)z)-\exp(xz)}{\exp(z)-1}\\
= m! [z^m] \exp(xz) \frac{\exp(z)-1}{\exp(z)-1}
\\ = m! [z^m] \exp(xz) = x^m.$$
For the second part we obtain that
$$\sum_{k=0}^n (-k)^m = \sum_{k=0}^n\left( S_m(1-k)-S_m(-k)\right)
= S_m(1)-S_m(-n)$$
because the terms telescope. But
$$S_m(1) = m! [z^m] \frac{\exp(1\times z)-1}{\exp(z)-1}
= m! [z^m] 1 = 0,$$
so the result follows.
