The solution to this problem is probably painfully obvious to most people, but I'm not making the leap for some reason. The problem is:

So, by Fermat's Little Theorem, we know that:Show that $\displaystyle 1^p+2^p+3^p+...(p-1)^p \cong 0(mod\; p)$ when p is an odd prime.

$\displaystyle a^p \cong a(mod\; p)$ and $\displaystyle a^{p-1} \cong 1(mod\; p)$.

And thus, $\displaystyle 1^p+2^p+3^p+...(p-1)^p \cong 1+2+3+... (p-1)\;(mod\; p).$

So, by my thinking, I need to show that $\displaystyle 1+2+3+... (p-1) \cong 0\;(mod\; p)$, but I'm not sure how to proceed.

Any help at all would be greatly appreciated.