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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.

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    What is $n$???? – 2017-01-29
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    if n=p here (otherwise i can't make sense of this) , we have: 1+2+..+p-1=p*(p-1)/2 Hence we want: K integer, (p-1)((p-2)! -1)=K*p*(p-1)/2 <=>2[(p-2)! -1]=K*p So it is equivalent (thanks to Gauss's theorem) to : p divides [(p-2)! -1] I believe , you should focus on that – 2017-01-29

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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 $$(p-1)! \equiv (p-1) (\bmod \frac{p(p-1)}{2}).$$

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    Yes, the n is supposed to be a p, sorry for the typo! – 2017-01-29
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    Thank you for the guidance! – 2017-01-29