Sum of the values of a polynomial over a finite field

You just need to show that $\sum_{\alpha\in F}\alpha^k=0$ for $k=0,1,\dots,q-2$. This is clear for $k=0$ (understanding $0^0$ as $1$). But $\alpha^q-\alpha=0$ for all $\alpha$ so $\alpha^{q-1}-1=0$ for all $\alpha\ne0$, and the result follows from the Newton identities.

Nothing wrong with Andres' answer (+1). Offering a possibly technically simpler approach based on the cyclicity of the multiplicative group $F^*=GF(q)\setminus\{0\}$. We know that the group consists of powers of a generator (aka a primitive element) $g$ of order $q-1$. Thus $F^*=\{1,g,g^2,g^3,\ldots,g^{q-2}\}.$

So if $k$ is an exponent, $0< k\le q-2$, then $$ \sum_{\alpha\in F}\alpha^k=\sum_{\alpha\in F^*}\alpha^k=\sum_{i=0}^{q-2}g^{ki}=\frac{1-g^{(q-1)k}}{1-g^k}=\frac{1-1}{1-g^k}=0 $$ by the formula for a geometric sum.

OTOH if $k=0$, then $$ \sum_{\alpha\in F}\alpha^0=\sum_{\alpha\in F}1=q\cdot 1=0. $$

The claim follows by taking a linear combination of sums like these.

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This result comes in handy in coding theory. For example in the study of the duals of Reed-Solomon codes. It also shows up at other places such as in this question, where it is used as a tool of discrete $GF(q)$-valued Fourier analysis of functions $F^*\to F^*$.

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EDIT:

An even simpler way is the following. Let use fix an exponent $k$, $0<k<q-1$. The polynomial $x^k-1$ is of degree $k$, so it has at most $k$ zeros in $GF(q)$. Thus there exists a non-zero element $\beta\in F^*$ such that $\beta^k\neq1$. The summation variable $\alpha$ ranges over $GF(q)$ while $\beta\alpha$ does, so the sum $$ S(k)=\sum_{\alpha\in GF(q)}\alpha^k=\sum_{\beta\alpha\in GF(q)}\beta^k\alpha^k= \sum_{\alpha\in GF(q)}\beta^k\alpha^k=\beta^kS(k). $$ As $\beta^k\neq1$ this equation implies that $S(k)=0$.