TITLE: Sums of powers below a prime
QUESTION [10 upvotes]: Given a prime $p$ and a natural number $k$, such that $k$ is not divisible by $p - 1$, prove that $\sum_{i = 1}^{p - 1}i^k \equiv 0 \pmod p$.
I split the proof into two cases: one where $k$ is odd and another where it is even.
The case where $k$ is odd can be proven as follows:
$$\sum_{i = 1}^{p - 1}i^k \equiv \sum_{i = 1}^{\frac{p - 1}{2}}i^k + \sum_{i = 1}^{\frac{p - 1}{2}}(-i)^k \pmod p$$
Since $k$ is odd each $(-i)^k = -i^k$, and will cancel with the positive $i$ leaving a zero. Therefore, 
$$\sum_{i = 1}^{p - 1}i^k \equiv 0 \pmod p \text{ if $k$ is odd.} $$ 
I approached the case where $k$ is even in a similar fashion. 
$$\sum_{i = 1}^{p - 1}i^k \equiv \sum_{i = 1}^{\frac{p - 1}{2}}i^k + \sum_{i = 1}^{\frac{p - 1}{2}}(-i)^k \equiv 2(\sum_{i = 1}^{\frac{p - 1}{2}}i^k) \space \pmod p$$
Since $p$ is prime, we just need to prove that $\displaystyle\sum_{i = 1}^{\frac{p - 1}{2}}i^k \equiv 0 \pmod p$.
I was stumped after this, so I considered a counterexample, one where $k$ is divisible by $p - 1$. In that I considered a case where $p = 5, k = 8$:
$$1^8 + 2^8 + 3^8 + 4^8 \equiv 1 + 1 + 1 + 1 \space \pmod p$$
This and a few other cases led to the conjecture that all the numbers below $p$ have a mod cycle $(mod \space p)$ that is divisible by $p - 1$. I tested it with $p = 7$.
$2: 2, 4, {\color{red}1}$
$3: 3, 2, 6, 4, 5, {\color{red}1}$
$4: 4, 2, {\color{red}1}$
$5: 5, 4, 6, 2, 3, {\color{red}1}$  
After this I was stuck. I couldn't see how this would help me solve the problem. And, more importantly, I couldn't prove my conjecture.

REPLY [4 votes]: All nonzero numbers less than p are roots of $x^{p-1}=1$ (mod p). Thus, all the elementary symmetric polynomials in these number of order less than p-1 are zero (mod p), so any symmetric polynomial in these numbers of order less than p-1 must be zero (mod p).