How to prove $\sum\limits_{k=0}^n{n \choose k}(k-1)^k(n-k+1)^{n-k-1}= n^n$? How do I prove the following identity directly?
$$\sum_{k=0}^n{n \choose k}(k-1)^k(n-k+1)^{n-k-1}= n^n$$
I thought about using the binomial theorem for $(x+a)^n$, but got stuck, because I realized that my $x$ and $a$ in this case are dynamic variables. Any hints? Don't give me the answer; I really want to think through this on my own, but a nudge in the correct direction would be awesome. Thanks! 

A nice proof uses the Lagrange-Bürman inversion formula. Start defining:
\begin{equation}
C(z) = z e^{C(z)}
\end{equation}
which gives the expansion:
\begin{equation}
e^{\alpha C(z)} = \alpha \sum_{n \ge 0} \frac{(\alpha + n)^{n - 1} z^n}{n!}
\end{equation}
Then you have:
\begin{equation}
e^{(\alpha + \beta) C(z)} = e^{\alpha C(z)} \cdot e^{\beta C(z)}
\end{equation}
Expanding and comparing both sides gives Abel's binomial theorem:
  \begin{equation}
    (\alpha + \beta) (\alpha + \beta + n)^{n - 1}
      = \sum_{0 \le k \le n}
   \binom{n}{k}
     (\alpha + k)^{k - 1}
     (\beta + n - k)^{n - k - 1}
  \end{equation}
