Proving Combinatorical Summation: $n!=\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^n$ been stuck with this question for the last few hours, any help would be appreciated.
$$
{\large n! = \sum_{k = 0}^{n}\left(-1\right)^{k}{\,n\, \choose \,k\,}
\left(\,n - k\,\right)^{n}}
$$
what I did:
$\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^n=n!\sum_{k=0}^n\frac{(-1)^k(n-k)^n}{k!(n-k)!}.$
So we are left to prove $\sum_{k=0}^n\frac{(-1)^k(n-k)^n}{k!(n-k)!}=1$. tried doing so using induction, or treating the sum as geometric sequence (which turned out poorly)
Suggestions?

First, ask yourself the following question: $\quad\displaystyle\sum_{k=0}^na^k{n\choose k}x^{n-k}~=~?\quad$ Hint: See binomial theorem.
Then apply the following two steps repeatedly: $(1).$ Differentiate both sides with respect to x, and 
$(2).$ Multiply both sides with x. Notice how, after each two steps, $\bigg(x\dfrac d{dx}\bigg)^k\circ(x-1)^n$ can be 
rewritten as $(x-1)^{n-k}\cdot P_k(x)$, where $P_k(x)$ is a polynomial of degree k in x and n. Notice also 
that $P_k(1)=n(n-1)\cdots(n-k+1)$. Try to prove these two observations by induction. Then it 
is quite evident that $P_n(1)$ will be $n!$
