I wish to show that $(p-2)! \equiv 1 \pmod p$ for a prime $p \ge 3$ using the fact that $(p-1)! \equiv -1 \pmod p$. (Deducing the latter is a later, and more advanced task.)
We have that $(p-1)(p-2)! \equiv -1 \pmod p$. We have that $(p-1) \equiv (p-1)\pmod p$, but this does not help. A weird thought along the lines of "What if I went to the other way around?" popped in my head, I wrote $(p-1) \equiv -1 \pmod p$ and $(-1)(p-2)! \equiv -1 \pmod p \implies (p-2)! \equiv 1 \pmod p$.
Much to my surprise, this was the right answer. I don't understand much of it, so this is a classic example of the right answer for the wrong reasons. If my methods are indeed valid; why do they work? If they are not; how would one normally go about solving this task, and how does one validate each step, basing ones arguments in elementary arithmetic?
The focus of this answer would be on its explanation rather than its content.
Notice that for all prime (integers, more generally) $p \ge 3$, we have
$$(p - 1)! = (p-1)(p-2)!$$
Now, suppose $(p-1)! \equiv -1 \pmod{p}$. This is the assumption in the question. It is the consideration we have to make. Then, taking both sides modulo $p$, we have
$$-1 \equiv (p-1)(p-2)! \pmod p$$
But $p -1$ gives a remainder of $-1$ when divided by $p$. In other words, $$p-1\equiv -1\pmod p$$
Substituting back, we have:
$$-1\equiv -1\cdot(p-2)!\pmod p$$
Multiplying both sides by $-1$, we have
$$(p-2)! \equiv 1 \pmod p$$
What this says is that
$$(p-1)!\equiv -1 \pmod p \implies (p-2)! \equiv 1 \pmod p$$
for all prime $p \ge 3$.
Now, I believe you would have to prove $p \ge 3$ prime $\implies (p-1)! \equiv -1 \pmod p$, to prove that the above deduction is true for $p \ge 3$ prime.