 Solution

For $n\ge 1$, the zero residue contributes zero. Multiplication by any nonzero $\alpha$ permutes the nonzero residues, so the sum $S$ satisfies $S={\alpha }^{n}S$ in ${\mathbb{F}}_p$. If $p-1\nmid n$, choose a primitive root $\alpha$; then ${\alpha }^n\ne 1$, forcing $S=0$. If $p-1\mid n$, every nonzero term equals one, so $S=p-1=-1$ in the field.

The printed display also mentions $n=0$, although its introductory assumption is $n\ge 1$. For that extra case the sum of the $p$ integer values $x^0$ for $1\le x\le p$ is $p\equiv 0$, so that branch is consistent too. The result is