TITLE: Expressing a factorial as difference of powers: $\sum_{r=0}^{n}\binom{n}{r}(-1)^r(l-r)^n=n!$?
QUESTION [16 upvotes]: 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.

REPLY [4 votes]: In the following we use the coefficient of operator $[z^r]$ to denote the coefficient of $z^r$ in a series. This way we can write e.g.
\begin{align*}
\binom{n}{r}=[z^r](1+z)^n\qquad\text{and}\qquad  r^n=n![z^n]e^{rz}
\end{align*}

We obtain for $n\geq 0$ and $l\in\mathbb{C}$
  \begin{align*}
\sum_{r=0}^n&\binom{n}{r}(-1)^r(l-r)^n\\
&=\sum_{r=0}^\infty[z^r](1+z)^n(-1)^rn![u^n]e^{(l-r)u}\tag{1}\\
&=n![u^n]e^{lu}\sum_{r=0}^\infty\left(-e^{-u}\right)^r[z^r](1+z)^n\tag{2}\\
&=n![u^n]e^{lu}\left(1-e^{-u}\right)^n\tag{3}\\
&=n![u^n]u^n\tag{4}\\
&=n!\\
\end{align*}
  and the claim follows.
Please note the formula is valid for any $l$.

Comment:

In (1) we apply the coefficient of operator to $\binom{n}{r}$ and to $(l-r)^n$ using $e^{(l-r)u}$. We also extend the limit to infty without changing anything since we are adding zeros only.
In (2) we do some rearrangements using the linearity of the coefficient of operator.
In (3) we apply the substitution rule
\begin{align*}
A(u)=\sum_{r=0}^\infty a_r u^r=\sum_{r=0}^\infty u^r [z^r]A(z)
\end{align*}
In (4) we consider  the series expansion
\begin{align*}
e^{lu}(1-e^{-u})^n&=(1+lu+\cdots)\left(u-\frac{u^2}{2}+\cdots\right)^n\\
&=1\cdot\left(u^n\mp\text{powers of }u\text{ greater than }n\cdots\right)
\end{align*}
and need only to respect the $u$-term with power equal to $n$.