How to prove $\sum_{i=1}^ki^k(-1)^{k-i}\binom {k+1}{i} =(k+1)^k$

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.

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.