TITLE: Prove $\sum_{k=0}^{n}\frac{n!}{k!}(n-k)n^k=n^{n+1}$ for any $n\in\mathbb N$.
QUESTION [8 upvotes]: I want to prove the following:
$$\sum_{k=0}^{n}\frac{n!}{k!}(n-k)n^k=n^{n+1}\quad\text{for any $n\in\mathbb N$.}$$ I tried induction and invoking the binomial theorem, to little avail. I’m looking for some quick and dirty solution. Thanks for any hints.

As the answers below reveal, the following update I had added earlier is not really of much use, so I struck it.
Update: After some rearrangements, the left-hand side above can be rewritten as $$\require{enclose}
     \enclose{horizontalstrike}{n\sum_{k=0}^{n-1}\binom{n-1}{k}n^k(n-k)!}$$ This form seems to suggest resorting to the binomial theorem.

REPLY [3 votes]: Here’s a combinatorial argument.
Clearly $n^{n+1}$ is the number of functions from $\{0,1,\ldots,n\}$ to $[n]$. If $f$ is any such function, let
$$k_f=\min\left\{k\in[n]:\exists\ell<k\big(f(k)=f(\ell)\big)\right\}\;.$$
For a given $k\in[n]$, how many of these functions have $k_f=k$?

There are $n^{\underline k}=n(n-1)\ldots(n-k+1)$ ways to choose $k$ distinct values for $f(0),\ldots,f(k-1)$.  
There are then $k$ ways to choose one of these values for $f(k)$.  
And finally there are $n^{n-k}$ ways to choose values for $f(i)$ with $k<i\le n$.

Summing over the possible values of $k$ then yields the result:
$$\begin{align*}
n^{n+1}&=\sum_{k=1}^nkn^{\underline k}n^{n-k}\\
&=\sum_{k=0}^nkn^{\underline k}n^{n-k}\\
&=\sum_{k=0}^n(n-k)n^{\underline{n-k}}n^k\\
&=\sum_{k=0}^n\frac{n!}{k!}(n-k)n^k\;.
\end{align*}$$