Let $q:=p^n$ with $p$ a prime and let $\mathbb{F}_q$ denote a finite field with $q$ elements. Consider the following result: If $f\in\mathbb{F}_q[X]$ has $\deg f\leq q-2$ then $$\sum_{x\in \mathbb{F}_q} f(x)=0.$$
This is called by Terry Tao the mean zero condition in Some notes on "non-classical" polynomials in finite characteristic, Corollary 2 (he considers the case $q=p$ only), and is the keystone of the short proof of the Chevalley-Warning theorem found in Wikipedia.
I have thought of a proof: It is enough to show the result for $f(X)=X^k$ with $0\leq k\leq q-2$, and $\sum_{x\in \mathbb{F}_q} x^k=0$ follows from the Newton-Girard formulas for power sums in terms of the elementary symmetric polynomials $e_i$, and the fact that the elements $x\in\mathbb{F}_q$ are the roots of the polynomial $X^q-X$, so $e_i\neq0$ iff $i=q-1$ (in fact, with this method we can compute, by recursion, the actual value of $\sum_{x\in \mathbb{F}_q} f(x)$ for $f$ of any degree).
The proof by Tao, for $q=p$, uses linear algebra: Let $V_d$ denote the space of polynomials of degree $d$ or less, and define the linear map $\Delta:V_d\rightarrow V_{d-1}$ such that $\Delta(f):=f(X+1)-f(X)$. It has $\ker\Delta=\langle 1\rangle$ while $d\leq p-1$. Since $\dim V_d=d+1$, this implies that $\Delta$ is surjective, so for every $f\in V_d$, $d\leq p-2$, there exists $g\in V_{d+1}$ such that $f=\Delta g$. Therefore $$\sum_{x\in\mathbb{F}_q} f(x)=\sum_{x\in\mathbb{F}_q} \Delta g(x)= \sum_{x\in\mathbb{F}_q} g(x+1)-g(x)=\sum_{x\in\mathbb{F}_q} g(x)-\sum_{x\in\mathbb{F}_q} g(x)=0.$$
My questions are:
1) Is there some way to extend the linear algebra proof to $\mathbb{F}_q$ so that the cases with $\deg f\geq p-1$ do not constitute a problem? (Note that $\Delta(X^p)=1=\Delta(X)$). My approach shows that a proof "without distinguished degrees" is possible.
2) Has this result a better-known name?
I'm also interested in other methods of proof (links would be nice).
