If P is prime, prove that $(p-1)!$ is congruent to $(p-1)$ $\pmod {1+2+3...+(p-1)}$ I have this as a problem to solve, but I'm not sure I'm headed in the right direction.
I tried to simplify a bit and got $(p-1)(1-(p-2))$ and use the definition of congruence to show $1+2+3...+n-1$ divides $(1-(p-2)!)$
I hit a dead end with that though and am not sure how to proceed.

The $n$ should be $p$ I suppose. Use the following facts:
$1 + 2 + \cdots + (p-1) = \frac{p(p-1)}{2}$ and $ (p-1)! \equiv -1 (\bmod p)$ and $(p-1)! \equiv 0 (\bmod (p-1)/2) $
$p$ and $\frac{p-1}{2}$ are relatively primes.
$ (p-1)! \equiv (p-1) (\bmod p)$ (Wilson's Theorem) and $(p-1)! \equiv (p-1) (\bmod (p-1)/2) $ gives that $$(p-1)! \equiv (p-1)  (\bmod \frac{p(p-1)}{2}).$$
