# $h+k=p-1$, $p$ prime. Prove $h!k! + (-1)^h \equiv 0 \pmod{p}$?

Suppose that $p$ is a prime. Suppose further that $h$ and $k$ are non-negative integers such that $h + k = p − 1$.

I want to prove that $h!k! + (−1)^h \equiv 0 \pmod{p}$

My first thought is that by Wilson's theorem, $(p-1)! \equiv -1 \pmod{p}$, and $h!k!$ divides $(p-1)!$ (definition of a binomial). Where would I go from here?

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$$h! = (-1)^h (p-1)(p-2) \dots (p-h) \mod p$$
$$\rm (p-1-k)\:!\ \equiv\ \frac{(-1)^{k+1}}{k!}\ \ (mod\ p),\ \ for\ p\ prime$$