TITLE: Sum of powers mod p
QUESTION [5 upvotes]: I've this problem that I did halve of the proof but I can't do the rest of it.

Let $p$ be an odd prime. We define $S_n$ as $S_n =  1^n +2^n + ... +(p-1)^n$
  Prove that
  $S_n \equiv
\begin{cases}
0  & \text{if $p-1 \nmid n $  } \\
-1, & \text{if $p-1 \mid n$ }
\end{cases}$

Partial proof:  
If $p-1\mid n$, then $\varphi(n)=p-1=kn$ for some k. 
As $p$ is prime, then every number from 1 to $p-1$ is relatively prime to it. So, using the Fermat-Euler Theroem, $a^{\varphi(n)} \equiv 1$ if $1 \leq a \leq p-1$. 
We see that every term in the sum becomes congruent to 1, and $\sum_1^{p-1} 1 = p-1 \equiv -1 \pmod p $.  
I don't know what to do if $p-1 \nmid n $. I've tried taking a primitive root mod p, but I got stuck.Any help or tip would be much aprecciated.

REPLY [5 votes]: To elaborate on my comment:  
Suppose that $p-1\nmid n$.  Then let $g$ be a primitive root $\pmod p$.  It follows that $g^n\not \equiv 1 \pmod p$.  Also, $g$ is clearly invertible $\pmod p$.  That implies that the list $\{g\times 1, g\times 2, \cdots, g\times (p-1)\}$ is a full list of the non-zero residues $\pmod p$.
Thus $$S_n\equiv (g\times 1)^n+(g\times 2)^n+\cdots +(g\times (p-1))^n\equiv g^nS_n \pmod p$$
But then, as $g^n\not \equiv 1$ we see that $S_n\equiv 0\pmod p$ as desired.