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.

In this answer are three different proofs of this cancellation lemma:
$$
\sum_{j=k}^n(-1)^{n-j}\binom{n}{j}\binom{j}{k}
=\left\{\begin{array}{}
1&\text{if }n=k\\
0&\text{otherwise}
\end{array}\right.\tag{1}
$$
Inductively, we can see that any polynomial in $j$ of degree $m$ can be written uniquely as
$$
\sum_{k=0}^mc_k\binom{j}{k}\tag{2}
$$
where the coefficient of $j^m$ is $\dfrac{c_m}{m!}$.
Putting $(1)$ and $(2)$ together, for $m\le n$, we have
$$
\begin{align}
\sum_{j=0}^{n}(-1)^{n-j}\binom{n}{j}\sum_{k=0}^mc_k\binom{j}{k}
&=\sum_{k=0}^m\sum_{j=0}^{n}(-1)^{n-j}\binom{n}{j}\binom{j}{k}\\
&=c_n\tag{3}
\end{align}
$$
If $m\lt n$, then $c_n=0$. If $m=n$, then $c_n$ is $n!$ times the coefficient of $j^n$.

Applying $(3)$
$$
\begin{align}
\sum_{r=0}^n\binom{n}{r}(-1)^r(l-r)^n
&=\sum_{r=0}^n(-1)^{n-r}\binom{n}{r}(r-l)^n\\
&=n!\tag{4}
\end{align}
$$
because, for any $l$, $(r-l)^n$ is a degree $n$ polynomial in $r$ in which the coefficient of $r^n$ is $1$.
