Prove that for any $a \in \mathbb{R}$
$$\sum_{k=0}^n (-1)^{k}\binom{n}{k}(a-k)^{n}=n!$$
I rewrote the sum as
$$\sum_{k=0}^n \left((-1)^{k}\binom{n}{k} \sum_{i=0}^n (-1)^{i}a^{n-i} k^{i} \right)$$
and then interchanged the summations, but that led me nowhere.
Any help will be appreciated.
Thanks!
Suppose that $a\ge n$ is an integer. For $k\in[n]$ let $A_k$ be the set of functions from $[n]$ to $[a]$ that do not have $k$ in their ranges. Then by a straightforward inclusion-exclusion argument
$$\sum_{k=0}^n(-1)^k\binom{n}k(a-k)^n\tag{1}$$
is the number of functions from $[n]$ to $[a]$ whose ranges include (and therefore are equal to) $[n]$. This, of course, is $n!$. Thus, $(1)$ is a polynomial function of the real variable $a$ that is equal to $n!$ for every integer $a\ge n$., so $(1)$ must be constant, and therefore
$$\sum_{k=0}^n(-1)^k\binom{n}k(a-k)^n=n!$$
for all $a\in\Bbb R$.
More generally if $p$ is any polynomial function of degree (at most)$~n$ with coefficient $c$ in degree$~n$, one has $$ \sum_{k=0}^n(-1)^k\binom nk p(x-k) = cn!, $$ independently of $x$. This applies here for $p:x\mapsto x^n$ with $c=1$. It easily follows by induction, given that $$ \sum_{k=0}^n(-1)^k\binom nk p(x-k) = \sum_{k=0}^{n-1}\binom{n-1}k (\Delta p)(x-k) \qquad\text{where $\Delta p:x\mapsto p(x)-p(x-1)$,} $$ because $\Delta p$ is a polynomial function of degree (at most)$~n-1$ with coefficient $nc$ in degree$~n-1$.
