 Solution

Split the factorial at $h$, using $k=p-1-h$:By Wilson theorem, this is $-1$. Therefore $h!k!\equiv (-1{)}^{k+1}(\mathrm{mod}p)$. If the prime number $p$ is odd, then $h+k=p-1$ is even, so $(-1{)}^k=(-1{)}^h$ andFor $p=2$, the signs $1$ and $-1$ are identical modulo $2$, and the two possibilities $(h,k)=(0,1),(1,0)$ also give the same result. The empty product convention $0!=1$ includes the endpoint cases. This is the complementary factorial congruence.