Given a prime number $p$ , establish the congruence: $(p-1)! \equiv (p-1) \pmod{1+2+3+\cdots+(p-1)}$
I have proceeded like this:
$\begin{align*}&(p-1)! \equiv (-1) \pmod{p} \quad \quad \quad \text{by Wilson's Theorem}\\ &(p-1)! \equiv 0 \pmod{\frac{p-1}{2}} \end{align*}$
Then I know that I have to apply Chinese remainder theorem but I don't have a thorough understanding of it.So please give me an elaborate answer to this question with respect to Chinese remainder theorem.