For p, an odd prime, prove that $(p-3)! ≡ (p-1)*(2)^{-1}\pmod p$ Here is what I got so far:
$$(p-3)! \equiv (p-1)(2)^{-1} \pmod p\\
(p-3)(p-2)(p-1)! \equiv (p-1)(2)^{-1}  \pmod p$$
Since $p$ is prime, by Wilson's theorem, $(p-1)! \equiv -1 \pmod p$. Hence,
$$\begin{align}(-1)(p-3)(p-2) &\equiv (p-1)(2)^{-1}\pmod p\\
-p^2 +5p -6 \equiv -6 &\equiv (p-1)(2)^{-1}\pmod p\\
-6(2) \equiv -12 &\equiv (p-1)\pmod p\\
-12 + 1 &\equiv p\pmod p\\
-11 &\equiv p\pmod p\end{align}$$
This is what I got so far. Am I doing this correctly? 
 A: Hint: We have 
$p-1\equiv -1\equiv (p-1)!=(p-3)!(p-1)(p-2)\equiv (p-3)!(-1)(-2)\equiv 2(p-3)!\pmod{p}$. 
Remark: Your proof begins in the right way, Wilson's Theorem is the correct tool.  After that there are problems. We are trying to get information about $(p-3)!$, but it does not appear explicitly in the rest of the calculation. 
A: In going from
$$(p-3)!\equiv (p-1)(2)^{-1}\pmod p$$
to
$$(p-3)(p-2)(p-1)!\equiv (p-1)(2)^{-1}\pmod p$$
you seem to be assuming that $(p-3)!=(p-3)(p-2)(p-1)!$.  That's not correct.  It works the other way:  $(p-1)!=(p-1)(p-2)(p-3)!$.  Consequently all the rest of your derivation is invalid, which you should have realized when you got to $-11\equiv p\pmod p$, which basically says that $p$ can only be $11$.
It is appropriate to apply Wilson's theorem, but you need to do it along lines like this:
$$\begin{align}
p-1&\equiv-1\\
&\equiv(p-1)!\qquad\text{(by Wilson's Theorem)}\\
&=(p-1)(p-2)(p-3)!\\
&\equiv(-1)(-2)(p-3)!\\
&=2(p-3)!\\
\end{align}$$
and then bring the $2$ to the other side as $2^{-1}$.
