How to use generating functions to prove that $n^n=\sum\limits^{n-1}_{k=0}\binom{n}{k}k^k(n-k)^{n-k-1}$

Question

Can anyone help? How can we use generating functions to prove the following identity:

$$n^n=\sum^{n-1}_{k=0}\binom{n}{k}k^k(n-k)^{n-k-1}$$

Answer 1

Take $\sum_{n=1}^{\infty}\frac{n^nx^n}{n!}$ as the exponential generating function of the left hand side.

$$\sum_{n=1}^{\infty}\sum_{k=0}^{n-1} \frac{{n \choose k}k^k(n-k)^{n-k-1}x^n}{n!}=\sum_{k=0}^{\infty} \frac{k^k}{k!}\left(\sum_{n=k+1}^{\infty}\frac{(n-k)^{n-k-1}x^n}{(n-k)!}\right)$$

$$\sum_{n=k+1}^{\infty}\frac{(n-k)^{n-k-1}x^n}{(n-k)!}=\sum_{n=1}^{\infty}\frac{n^{n-1}x^{n+k}}{n!}=-x^kW(-x)$$ where $W$ is the Lambert function, defined by $z=W(z)e^{W(z)}$ with the principal branch having the series expansion $W(x)=\sum_{n=1}^{\infty}\frac{(-n)^{n-1}x^n}{n!}$.

$$\sum_{n=0}^{\infty}\sum_{k=0}^{n-1} \frac{{n \choose k}k^k(n-k)^{n-k-1}x^n}{n!}=\sum_{k=0}^{\infty}\frac{k^k(-x^kW(-x))}{k!}=\left(\sum_{k=0}^{\infty}\frac{k^kx^k}{k!}\right)-1=\sum_{k=1}^{\infty}\frac{k^kx^k}{k!}$$