Let $\Bbb F$ be a field of size $q$, and let $u$ be a positive integer, such that $q-1$ does not divide $u$. Prove: $$\sum_{x\in \Bbb F}x^u=0$$
I thought about using the fact that $\Bbb F^*$ (the units in $\Bbb F$) is a cyclic group. I know I can rewrite $u=(q-1)k+r$, and then because $x^{q-1}=1$, the problem reduces to calculating $\sum_{x\in \Bbb F}x^r$, $0<r<q-1$. I'm pretty stuck from here. Any ideas?
Hint:
For some non zero $y\in\Bbb F$, such that $y^u\neq 1$, show that $$y^u\sum_{x\in\Bbb F}x^u=\sum_{x\in\Bbb F}x^u\tag{1}$$
Edit for completion:
The map $\phi:\Bbb F\rightarrow\Bbb F$ given by $\phi(x)=yx$ is a non-trivial group automorphism if $y\neq 0,1$. This means that we have $$\{x\in\Bbb F\}=\{yx\mid x\in\Bbb F\}\tag{2}$$ consequently $$\sum_{x\in\Bbb F}x^u=\sum_{x\in\Bbb F}(yx)^u\tag{3}$$ As $y,u$ are fixed, we can rewrite $(3)$ as $$\left(1-y^u\right)\sum_{x\in\Bbb F}x^u=0$$ Since we're working over a field, and $y\neq 0,1$, it follows that $$\sum_{x\in\Bbb F}x^u=0\tag{4}$$