Prove $\sum_{k=0}^{n}\frac{n!}{k!}(n-k)n^k=n^{n+1}$ for any $n\in\mathbb N$.

One way is to rewrite the L.H.S. as a telescoping sum: $\displaystyle\sum_{k=0}^{n}\left(\frac{n!~n^{k+1}}{k!}-\frac{n!~n^k}{(k-1)!}\right)$.

Can you show $$\sum_{k=0}^m\frac{n!}{k!}(n-k)n^k=\frac{n!}{m!}n^{m+1}$$

Suppose we seek to evaluate

$$\sum_{k=0}^n \frac{n!}{k!} (n-k) n^k = n! n^n \sum_{k=0}^n \frac{n-k}{k!} n^{k-n}.$$

Introduce

$$n^{k-n} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{z^k}{z^{n+1}} \frac{1}{1-z/n} \; dz.$$

Observe that this integral provides an Iverson bracket, as it vanishes when $k\gt n.$ Therefore we may extend $k$ to infinity.

We get for the sum

$$n! n^n \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{1-z/n} \sum_{k\ge 0} \frac{n-k}{k!} z^k \; dz \\ = n! n^n \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{1-z/n} \left(n \exp(z) - z \sum_{k\ge 1} \frac{1}{(k-1)!} z^{k-1}\right) \; dz \\ = n! n^n \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{n}{n-z} \left(n \exp(z) - z \exp(z)\right) \; dz \\ = n! n^{n+1} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \exp(z) \; dz = n! n^{n+1} \frac{1}{n!} = n^{n+1}.$$

This concludes the argument.