Given an odd prime $p$, prove that $2(p - 3)!$ is congruent to $-1 \bmod p$ My question is:

$p$ is odd and prime, prove that $2(p - 3)!$ is congruent to  $-1 \bmod p.$

Here is what I have so far:
$ (p - 1)! = (p - 1)(p - 2)(p - 3)!$ 
$(p - 1)(p - 2)(p - 3)! \equiv -1 \bmod p$  
I know I need to get rid of $p - 1$ and $p - 2$ but I do not know how to find their inverses.
Any help is appreciated. Thank you!  
 A: So you know that $(p - 1)! \equiv -1 \pmod{p}$ -- this is Wilson's theorem. Expanding that out,
$$ (p - 1) \times (p - 2) \times (p - 3)! \equiv -1 \pmod{p}.$$
But since you're working mod $p$, you can replace $p - 1$ with $-1$ in this product. since they are congruent mod $p$. Similarly you can replace $p - 2$ with $-2$.
Edited to add: Why can we replace $p - 1$ with $-1$?  This is one of the laws of modular arithmetic. This appears to be a homework problem, so I'm assuming you have a textbook that at least states these rules (perhaps without proof). The general rule here can be expressed as:

For integers $a, b, c$ and positive integer $n$, if $a \equiv b {\pmod n}$, then $ac \equiv bc {\pmod n}$.

Why is this true, in general?  If $a \equiv b {\pmod n}$, then by definition $b - a$ is a multiple of $n$ - call it $kn$ where $k$ is some integer.  But then $bc - ac = c(b - a)$ is a multiple of $n$, namely $ckn$.  So $ac \equiv bc {\pmod n}$ by the definition of congruence.
So now let's go back to specific case you're asking about.  Let $a = p - 1$, $b = -1$, $c = (p - 2) \times (p - 3)!$, and $n = p$.  It's obvious that $(p - 1) \equiv -1 {\pmod n}$ and so you get
$$(p - 1) \times (p - 2) \times (p - 3)! \equiv -1 \times (p - 2) \times (p - 3)! \pmod{p}$$
which justifies replacing the $p - 1$ with $-1$.   Repeating this justifies the replacement of $p - 2$ with $-2$.
