Let $p$ be prime and $\gcd (n, p) = 1$.
Prove that
$$ (p− 1)! + n^{p−1} \equiv 0 \pmod p $$
We're given $p$ is prime. By Wilson's Theorem this means
$$ (p-1)! \equiv -1 \pmod p $$
Also, $\gcd(n,p)=1$ means $p$ does not divide $n$, and so Fermat's Little Theorem tell us
$$ n^{p-1} \equiv 1 \pmod p$$
Adding these gives us the desired result.
$$ (p− 1)! + n^{p−1} \equiv -1 + 1 \equiv 0 \pmod p $$