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 answer is similar to Moron's but I think is somewhat simpler.  Expand $(e^x-1)^n$ using the binomial theorem to obtain $$(e^x-1)^n = \sum_{j=0}^n (-1)^{n-j} \binom{n}{j} e^{jx}.$$  Differentiate both sides $n$ times.  Every term on the left-hand side will contain an $e^x-1$ factor except for an $n!e^{nx}$ term, and the right side will be $$\sum_{j=0}^n (-1)^{n-j} \binom{n}{j} j^n e^{jx}.$$  Then substitute $x = 0$ to obtain 
$$n! = \sum_{j=0}^n (-1)^{n-j} \binom{n}{j} j^n.$$  Reindex the sum and divide by $n^n$.  
