My attempt: From Wilson's Theorem: For a prime $p$,
$$(p-1)! \equiv (-1) \pmod p$$
Multiplying both sides by $(p-2)$,
$$(p-2)! \equiv -(p-2) \pmod p$$ i.e. $$(p-2)! \equiv 2 \pmod p$$
So the remainder is 2. But the answer is given to be 1. So where am I wrong?
EDIT: after being pointed out where I was wrong in the comment (THANK YOU!), I see:
$$(p-1)(p-2)! \equiv (-1) \pmod p$$ i.e. $$(-1)(p-2)! \equiv (-1) \pmod p$$ i.e. $$(p-2)! \equiv (1) \pmod p$$
So the remainder is 1. Thank you! :)