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.

Lemma: Let $f(j) = \sum_{k=0}^n f_k j^k$ be a degree $n$ polynomial. I claim that $\sum_{j=0}^{n} (-1)^j \binom{n}{j} f(j) = n! f_n$. 
Proof: For any polynomial $g$, define $\Delta(g)$ to be the polynomial $\Delta(g)(j) = g(j) - g(j+1)$. Observe that, if $g$ is a polynomial with leading term $a x^d + \cdots$, then $\Delta(g)$ is a polynomial with leading term $-d a x^{d-1} + \cdots$. In particular, if $f$ is as in the statement of the lemma, then $\Delta^n(f)$ is the constant $(-1)^n n! f_n$. The sum in question is $\Delta^n(f)$ evaluated at $0$. QED
Now, apply the lemma to $f(j) = (1-j/n)^n$.
