Expressing a factorial as difference of powers: $\sum_{r=0}^{n}\binom{n}{r}(-1)^r(l-r)^n=n!$? The successive difference of powers of integers leads to factorial of that power. Here's the formula:

$$\sum_{r=0}^{n}\binom{n}{r}(-1)^r(n-r)^n=n!$$

Can anyone give a proof of this result?
Note: The original question was to prove the more general

$$\sum_{r=0}^{n}\binom{n}{r}(-1)^r(l-r)^n=n!$$

for an integer $l$, which some of the answers address.

This identity also has an algebraic proof. Suppose the sum we are trying to evaluate is given by
$$\sum_{k=0}^n {n\choose k} (-1)^k (n-k)^{n}.$$
Observe that when we multiply two exponential generating functions of the sequences $\{a_n\}$ and $\{b_n\}$ we get that
\begin{align}
 A(z) B(z) &= \sum_{n\ge 0} a_n \frac{z^n}{n!} \sum_{n\ge 0} b_n \frac{z^n}{n!}
= \sum_{n\ge 0} \sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\
&= \sum_{n\ge 0} \sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!}
= \sum_{n\ge 0} \left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}
\end{align}
i.e. the product of the two generating functions is the generating function of
$$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$
Now in the present case we clearly have
$$A(z) = \sum_{q\ge 0} (-1)^q \frac{z^q}{q!} = \exp(-z).$$
Furthermore we have
$$B_n(z) = \sum_{q\ge 1} q^{n} \frac{z^q}{q!}
 = \exp(z) \sum_{k=1}^{n} {n\brace k} z^k,$$
as can be seen by induction. For $n=1$ we have
$$B_1(z) = \sum_{q\ge 1} q \frac{z^q}{q!}
= z \sum_{q\ge 1} \frac{z^{q-1}}{(q-1)!} = \exp(z)\times  z.$$
Now using the induction hypothesis we have $$B_{n+1}(z) = z \frac{d}{dz} B_n(z)
= z \exp(z) \sum_{k=1}^{n} {n\brace k} k z^{k-1}
+ z \exp(z) \sum_{k=1}^{n} {n\brace k} z^k.$$
This is
\begin{align}
& z \exp(z) \sum_{k=0}^{n-1} {n\brace k+1} (k+1) z^k
+ z \exp(z) \sum_{k=1}^{n} {n\brace k} z^k \\
&= z \exp(z) \left({n\brace 1} +
\sum_{k=1}^{n-1} \left({n\brace k+1} (k+1)+{n\brace k}\right)z^k+
{n\brace n} z^{n}\right) \\ 
&=  z \exp(z) \left({n+1\brace 1} + \sum_{k=1}^{n-1} {n+1\brace k+1} z^k +
{n+1\brace n+1} z^{n}\right) \\ 
&= \exp(z) \left({n+1\brace 1} z + \sum_{k=1}^{n-1} {n+1\brace k+1} z^{k+1} +
{n+1\brace n+1} z^{n+1}\right)
= \exp(z) \sum_{k=1}^{n+1} {n+1\brace k} z^k.
\end{align}
Returning to the original problem we find that the sum has the value
$$n! [z^n] A(z) B_n(z) = n! [z^n] \sum_{k=1}^{n} {n\brace k} z^k
= n! {n\brace n} = n!$$
which was to be shown.
A useful exercise in the algebra of Stirling numbers and binomial coefficients. Here is a MSE link that points to another calculation of the same type.
Addendum. In response to the comment we can actually show that
$$\sum_{k=0}^n {n\choose k} (-1)^k (l-k)^{n}$$
for $l$ an integer.
Observe that $$l-k = (n-k)+(l-n)$$ and put $p=l-n.$
Using the same setup as before we now have
\begin{align}
B(z) &= \sum_{q\ge 0} (q+p)^n \frac{z^q}{q!}
= \sum_{q\ge 0} \sum_{m=0}^n {n\choose m} q^m p^{n-m} \frac{z^q}{q!}
\\
&= \sum_{q\ge 0} p^n  \frac{z^q}{q!} + 
\sum_{q\ge 0} \sum_{m=1}^n {n\choose m} q^m p^{n-m} \frac{z^q}{q!}
=  \exp(z) p^n +\sum_{m=1}^n {n\choose m} p^{n-m} \sum_{q\ge 0} q^m  \frac{z^q}{q!}
\\ 
&=  \exp(z) p^n +
\exp(z) \sum_{m=1}^n {n\choose m} p^{n-m}  \sum_{k=1}^m {m\brace k} z^k.
\end{align}
This formula also holds for $p=0$ if we assume that $0^0=1$, in which case it produces the first formula we derived above.
It follows that
$$n! [z^n] A(z) B_n(z) =
n! [z^n] \left(p^n+\sum_{m=1}^n {n\choose m} p^{n-m}  
\sum_{k=1}^m {m\brace k} z^k\right)
= n! {n\choose n} {n\brace n} = n!$$
thereby showing the result for all $l.$
Addendum Oct 10 2016. There is really nothing to prove here as the formula $$\frac{1}{n!} \sum_{k=0}^n {n\choose k} (-1)^{n-k} k^m$$ by definition gives the Stirling number of the second kind $${m\brace n}.$$
