Range of $m,n$ in $\sum_{j=0}^{m}{m\brace j}\binom{n}{j}j!=n^{m}$

Question

A well-known identity related to Stirling numbers of the second kind states:

$$\sum_{j=0}^{m}(-1)^j{m\brace j}j!=(-1)^m$$

I know the formula: $$\sum_{j=0}^{m}{m\brace j}\binom{n}{j}j!=n^{m}\tag{I}$$

But the combinatorial proof of this relation is valid for nonnegative integers $n,m$,however setting $-1 \mapsto n$and using negative binomial coefficient yields: $$\sum_{j=0}^{m}{m\brace j}\binom{-1}{j}j!=\sum_{j=0}^{m}(-1)^j{m\brace j}j!=(-1)^m$$

But how this is true?how are we allowed to left $n$ to be a negaive number?I guess if we want to use the relation $\text{(I)}$ then we first need to give another proof for the other $n,m$ which are not nonnegative integers,can someone explicitly determine the range of $m,n$?

Answer 1

Fix an integer $m\ge0$.

Your identity (I) is valid for all integers $n\ge0$. That's infinitely many.

If we write, for $x\in\Bbb R$ $$\binom{x}j=\frac1{j!}x(x-1)(x-2)\cdots(x-j+1)$$ then $$F_m(x)=\sum_{j=0}^m{m\brace j}\binom{x}{j}j!$$ is a polynomial of degree $m$. The identity $$F_m(x)=x^m\tag{*}$$ holds for all integers $x=n\ge0$. As it is true infinitely often, it is true for all real numbers, in particular for $x=-1$.