Proof of a combination identity:$\sum \limits_{j=0}^n{(-1)^j{{n}\choose{j}}\left(1-\frac{j}{n}\right)^n}=\frac{n!}{n^n}$ I want to ask if there is a slick way to prove:
$$\sum_{j=0}^n{(-1)^j{{n}\choose{j}}\left(1-\frac{j}{n}\right)^n}=\frac{n!}{n^n}$$
Edit:
I know Yuval has given a proof, but that one is not direct. I am requesting for a direct algebraic proof of this identity.
Thanks.

This is inclusion-exclusion for counting the number of onto maps from $n$ to $n$.
