A polynomial rep. for $\frac{1}{m!} \sum_{k=0}^n (-1)^{k+m} \frac{k^n}{k+1} \binom{m}{k}$ Let the generalized Bernoulli polynomials be defined by their generating function
$$  \sum_{n=0}^\infty \frac{B_n^{(s)}(x)}{n!} t^n = \Big(\frac{t}{e^t-1}\Big)^s\ e^{xt} $$
The Stirling numbers of the second kind can be written in terms of the Norlund polynomials $(B_n^{(s)}(0))$  as
$$ S_2(n,m)=\frac{1}{m!} \sum_{k=0}^n (-1)^{k+m} k^n \binom{m}{k} = \binom{n}{m} B_{n-m}^{\ (-m)}(0)$$
How can it be proved that the sum in the title has a generalized Bernoulli polynomial representation
$$ \frac{1}{m!} \sum_{k=0}^n (-1)^{k+m}  \frac{k^n}{k+1} \binom{m}{k}  
= \binom{n}{m+1} B_{n-(m+1)}^{(-(m+1))}(-1) + \frac{(-1)^{n+m}}{(m+1)!}  \quad ? $$
The proposed identity has been verified for many $(n,m),$ with n>m>0.  If you present a proof, is there a way to extend it to denominators other than $k+1$ within the summation on the left hand side?
 A: We seek to show that with $n\gt m\gt 0$
$$\frac{1}{m!} \sum_{k=0}^n (-1)^{k+m} \frac{k^n}{k+1}
{m\choose k}
\\ = {n\choose m+1} (n-m-1)! [t^{n-m-1}]
\frac{\exp(-t)}{t^{m+1}} (\exp(t)-1)^{m+1}
+ \frac{(-1)^{n+m}}{(m+1)!}.$$
The LHS is
$$\frac{1}{(m+1)!} \sum_{k=0}^n (-1)^{k+m} k^n
{m+1\choose k+1}.$$
We get for the coefficient extractor on the RHS
$${n\choose m+1} (n-m-1)! [t^n]
\sum_{k=0}^{m+1} {m+1\choose k} (-1)^{m+1-k}
\exp((k-1)t)
\\ = \frac{1}{(m+1)!}
\sum_{k=0}^{m+1} {m+1\choose k} (-1)^{m+1-k} (k-1)^n
\\ = \frac{1}{(m+1)!}
\sum_{k=-1}^{m} {m+1\choose k+1} (-1)^{m-k} k^n.$$
Restoring the extra term on the RHS yields
$$\frac{1}{(m+1)!}
\sum_{k=-1}^{m} {m+1\choose k+1} (-1)^{m-k} k^n
+ \frac{(-1)^{n+m}}{(m+1)!}
\\ = \frac{1}{(m+1)!}
\sum_{k=0}^{m} {m+1\choose k+1} (-1)^{m-k} k^n.$$
The LHS is the same as the RHS except for the upper limit of the sum
being $n$ instead of $m.$ Note however that ${m+1\choose k+1} =
(m+1)^{\underline{k+1}}/(k+1)! = 0$ when $k\gt m$ so we may lower the
limit  to $m$ (here we have used $n\gt m.$) This concludes the argument.
