 Solution

For $n\ge 1$, the zero residue contributes zero. Write $S={\sum }_{x\in {\mathbb{F}}_p^{\times }}{x}^n$. Multiplication by any nonzero $\alpha$ permutes these residues, so $S={\alpha }^{n}S$ in ${\mathbb{F}}_p$. If $(p-1)\nmid n$, choose a primitive root $g$; then $g^n\ne 1$, and $(1-g^n)S=0$ forces $S=0$.

If $(p-1)\mid n$, every nonzero $x$ has $x^n=1$, so $S=p-1\equiv -1(\mathrm{mod}p)$. ThereforeThe source also lists $n=0$, although its premise assumes $n\ge 1$. For that separate case, the integer sum has $p$ terms each equal to one, so it is zero modulo $p$.