Wilson's theorem, $(p-2)! \bmod p$ and $(p-3)! \bmod p$

Hint: Multiply by the multiplicative inverses of $p-1$ and $p-2$.

From Wilson theorem we have,

$(p-1)!=-1$ for $p$ prime. Now suppose $(p-2)!=x$, then $(p-1)!=(p-1)(p-2)!=(p-1) x=-1 \pmod p$.This implies $x=1\pmod p$.