modular arithmetic - When is $(p - 2)! equiv 1 (bmod p)$

I want to show when the following is true for $p$ a prime number. $(p - 2)! \equiv 1 \pmod p$. Could someone help me prove this? It worked for $p = 2$, $p = 3$, $p = 5$, so I believe it may work for all primes but I need to prove it. I don't know how to apply Wilson's or Fermat's theorem to this. I tried to rewrite it as $(p - 1 - 1)! \equiv 1 \pmod p$ but I still couldn't see how to apply Wilson's theorem to it. Could someone help me?