binomial identity with negatives Prove that 
$$\sum_{k=0}^n(-1)^k\binom{n+1}{k+1}(k+1)^n=0\;.$$
I tried finding a combinatorial interpretation but to no avail. Here is a combinatorial statement, however crappy. Suppose we have $n$ blanks on a sheet of paper. We choose $k$ (at least one) of them. Afterwards, we fill all blanks except for the leftmost one with an integer from $1$ to $k$. Show that the number of ways to do this with $k$ even is the same as the number of ways to do this with $k$ odd.
I don't see how it is the same with the identity posted in the comments. I did not introduce any factorials.
 A: This  sequence shows  up quite  frequently  here. Suppose  we seek  to
evaluate
$$\sum_{k=0}^n {n+1\choose k+1} (-1)^k (k+1)^q$$
whith $q$ a non-negative integer.
Introduce the integral representation
$$(k+1)^q
= \frac{q!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{q+1}} \exp((k+1)z) \; dz.$$
This gives for the sum
$$\frac{q!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{q+1}} 
\sum_{k=0}^n {n+1\choose k+1} (-1)^k \exp((k+1)z) \; dz
\\ = -\frac{q!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{q+1}} 
\sum_{k=0}^n {n+1\choose k+1} (-1)^{k+1} \exp((k+1)z) \; dz.$$
This is
$$-\frac{q!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{q+1}} 
\left((1-\exp(z))^{n+1}-1\right) \; dz$$
which evaluates to
$$-q! [z^q] (1-\exp(z))^{n+1}
\\ = -q! \times (n+1)! \times (-1)^{n+1}
[z^q] \frac{(\exp(z)-1)^{n+1}}{(n+1)!}.$$
This is
$$(n+1)! \times (-1)^{n}
\times {q\brace n+1}.$$
The above is zero when $q\le n$ which was to be shown.

Remark.  Here  we  have  used  the combinatorial  class  of  set
partitions which is given by
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SET}(\mathcal{U}\times\textsc{SET}_{\ge 1}(\mathcal{Z})),$$
producing the generating function
$$\exp(u(\exp(z)-1))$$
and in particular
$${n\brace k}
= n! [z^n] \frac{(\exp(z)-1)^k}{k!}.$$
A similar calculation can be found at this MSE link.
