A property of finite fields Why does the remark in Wiki's proof of Warning's theorem true?

If ${\displaystyle i<q-1}$ then ${\displaystyle \sum _{x\in \mathbb {F} }x^{i}=0}$

Best regards
 A: This a special case of something more general.
Fact 1. Suppose $F$ is a field, $G$ is a finite group, and $\varphi:G \to F^{\times}$ is a nontrivial group homomorphism. Then $\sum_{x\in G}\varphi(x)=0$.
Here:

*

*$0$ is the additive identity of $F$,

*$F^{\times}$ is the multiplicative group $F\setminus \{0\}$

*$\varphi$ nontrivial means there is some $a\in G$ such that $\varphi(a)\neq 1$, where $1$ the multiplicative identity of $F$.

Proof: Let $S=\sum_{x\in G}\varphi(x)$. Fix $a\in G$ such that $\varphi(a)\neq 1$. Then we have
$$
\varphi(a)S=\sum_{x\in G}\varphi(a)\varphi(x)=\sum_{x\in G}\varphi(ax)=S
$$
The last equality uses the fact that as $x$ ranges over $G$ so does $ax$ (with $a$ fixed). So we have $\varphi(a)S=S$, which means $S=0$ since $\varphi(a)\neq 1$. The proof is complete.
So you can apply this fact in your case to $G=\mathbb{F}^{\times}$. (We assume $q=|\mathbb{F}|$.) As noted by others, $x\mapsto x^i$ is a (nontrivial) homomorphism from $\mathbb{F}^{\times}$ to itself, and clearly adding $x=0$ to the sum does nothing.
The only subtle point is proving that $x\mapsto x^i$ is nontrivial. As others note, this follows from the fact that $\mathbb{F}^{\times}$ is cyclic, and so if $a$ is a generator then $a^i\neq 1$ for $i<q-1$. But you don't technically need this, you just need that if $i<q-1$ then there is some $a\in\mathbb{F}^{\times}$ such that $a^i\neq 1$. This follows from the fact that in any field, the polynomial $x^i-1$ has at most $i$ solutions. (Admittedly, that last fact is a big part of the usual proofs that $\mathbb{F}^{\times}$ is cyclic.)
A: The map $x \mapsto x^i$ is a group homomorphism $\mathbb{F}^{\times} \to \mathbb{F}^{\times}$.
If its image is a subgroup of $H$ order $d$, then the image is exactly the set of solutions of $x^d=1$ because $\mathbb{F}^{\times}$ is cyclic. Therefore, the sum of its elements is $0$ because there is no $x^{d-1}$ term in $x^d-1$. Finally, $$\sum _{x\in \mathbb {F} }x^{i}=\sum _{x\in \mathbb {F}^\times}x^{i}= m  \sum _{y \in H} y=0$$ because the preimage of every $y \in H$ has the same number of elements $m$, the order of the kernel.
