A summation identity over prime fields

Question

Let $p > 2$ be a prime. Can someone prove that for for any $t \leq p-2$ the following identity holds

$\displaystyle \sum_{x \in \mathbb{F}_p} x^t = 0$

Answer 1

Let $t\in\{1,\ldots,p-2\}$ and let $g\in\Bbb{F}_p^{\times}$ be a primitive root. Then $g^t\neq1$ and $g^{p-1}=1$, so $$\sum_{x\in\Bbb{F}_p}x^t=\sum_{k=0}^{p-2}g^{kt}=\frac{1-g^{(p-1)t}}{1-g^t}=0.$$

Answer 2

Because of $t < p-1$, there exists an $a\in\mathbb F_p^\times$ with $a^t\neq 1$. Now $$ \sum_{x\in\mathbb F_p} x^t = \sum_{x\in\mathbb F_p} (ax)^t = a^t \sum_{x\in\mathbb F_p} x^t. $$ So $$\underbrace{(1 - a^t)}_{\neq 0}\sum_{x\in\mathbb F_p} x^t = 0$$ and therefore $$\sum_{x\in\mathbb F_p} x^t = 0.$$