Sum of powers in finite fields

Today's proof is from A Course in Arithmetic by Serre. Here we'll let $0^0 = 1$.

Let $u \geq 0$, and let $K$ be a finite field of order $q = p^l$ (and hence of characteristic $p$). Then the sum
$$S^u(x) = \displaystyle\sum_{x \in K} x^u$$ is equal to $-1$ if $u \geq 1$ and divisible by $q - 1$; otherwise, the sum equals zero.

If $u = 0$, then we immediately get $S^u(x) = p^l \cdot 1 = 0$.

If $u \geq 1$ and is divisible by $q - 1$, we have $0^u = 0$ and $x^u = 1$ if $x \neq 0$. Therefore, $S^u(x) = (q - 1) \cdot 1 = -1$.

If $u \geq 1$ and not divisible by $q - 1$, then there exists $y \in K^\times$ such that $y ^u \neq 1$. (We can let $y$ be a generator of $K^\times$.) Then
$$S^u(x) = \displaystyle\sum_{x \in K^\times} x^u = \displaystyle\sum_{x \in K^\times} y^u x^u = y^uS^u(x),$$ so $S^u(x) = 0$.