TITLE: How to prove $\sum_{i=1}^ki^k(-1)^{k-i}\binom {k+1}{i} =(k+1)^k$
QUESTION [9 upvotes]: How to prove $\sum_{i=1}^ki^k(-1)^{k-i}\binom {k+1}{i} =(k+1)^k$
where k is a positive integer.
Any hints can help.

REPLY [2 votes]: Suppose we seek to verify that
$$\sum_{k=0}^n k^n (-1)^{n-k} {n+1\choose k} =
(n+1)^n.$$
Re-write this as
$$\sum_{k=0}^{n+1} k^n (-1)^{n-k} {n+1\choose k} = 0.$$
Introduce
$$k^n = 
\frac{n!}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n+1}} \exp(kz) \; dz.$$
This yields for the sum
$$\frac{n!}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n+1}} 
\sum_{k=0}^{n+1} (-1)^{n-k} {n+1\choose k}
\exp(kz) \; dz
\\ = \frac{n!}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n+1}} (\exp(z)-1)^{n+1} \; dz.$$
This is
$$[z^n] (\exp(z)-1)^{n+1} = 0$$
because $$\exp(z)-1 = z + \frac{z^2}{2} + \frac{z^3}{6} +\cdots$$
This is essentially the same as the answer by @MarkusScheuer which I upvoted.