How to prove $\sum_{i=1}^ki^k(-1)^{k-i}\binom {k+1}{i} =(k+1)^k$ 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.
 A: 
We can write OPs claim by putting the RHS to the left as:
\begin{align*}
  \sum_{i=1}^{k+1}\binom{k+1}{i}(-1)^{k-i}i^k=0\tag{1}
  \end{align*}

We use the  coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of a series.

We observe
  \begin{align*}
  [x^k](e^x-1)^{k+1}&=[x^k]\left(\sum_{j=1}^{\infty}\frac{x^j}{j!}\right)^{k+1}\\
  &=0\tag{2}
  \end{align*}
On the other hand we obtain
  \begin{align*}
  [x^k](e^x-1)^{k+1}&=[x^k]\sum_{j=0}^{k+1}\binom{k+1}{j}(-1)^{k+1-j}e^{jx}\\
  &=[x^k]\sum_{j=0}^{k+1}\binom{k+1}{j}(-1)^{k+1-j}\sum_{r=0}^{\infty}j^r\frac{x^r}{r!}\\
  &=\frac{1}{k!}\sum_{j=1}^{k+1}\binom{k+1}{j}(-1)^{k+1-j}j^k\tag{3}\\
  \end{align*}

Combining (2) and (3) and multiplying with $-\frac{1}{k!}$ we get
\begin{align*}
  \sum_{j=1}^{k+1}\binom{k+1}{j}(-1)^{k-j}j^k=0
  \end{align*}
and the claim follows.
A: 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.
