I thought of using Wilson's theorem for the proof.
First we have by Wilson's theorem $$(p - 1)!+1 \equiv 0 \pmod p$$ We can write this as $$(p - 2)!(p-1)+1 \equiv 0 \pmod p$$ $$(p - 2)!(p-1)+1=p(p-2)!-[(p-2)!-1]$$ Hence the right side p(p-2)! is divisible by p also the other term is divisible by p. So we have $$(p - 2)!-1 \equiv 0 \pmod p$$.
Is this correct?