 Solution

For $k=0$, the convention $0^0=1$ gives $S_0=p\equiv 0(\mathrm{mod}p)$. Now suppose $1\le k<p-1$ and choose a primitive root $g$ modulo $p$. The nonzero residues are $g^j$ for $0\le j\le p-2$, while the $x=0$ term vanishes. HenceThis is a finite geometric series. Since $g^k\not\equiv 1(\mathrm{mod}p)$ but $g^{k(p-1)}\equiv 1(\mathrm{mod}p)$ by Fermat's little theorem, multiplication by the nonzero residue $g^k-1$ givesThe factor $g^k-1$ is invertible modulo the prime number $p$, so