There is a result of Pierce (see here) which states:
Let $p\ge 3$ be a prime, and fix $n,k$ positive integers such that there exactly $2k$ non-zero $n$-th residues modulo $p$. Then $$ \sum (x_1\cdots x_k)^2 \equiv 2(-1)^{k-1} \bmod{p}, $$ where the sum is taken over all pairwise distinct $n$-residues $x_1,\ldots,x_k$.
In particular, if the order matters then the above result would be multiplied by a factor $k!$. Here, I would ask for a generalization. Is it possible to prove in a similar way the following one?
Question. Let $p\ge 3$ be a prime, and fix $n,k$ positive integers such that there exactly $2k$ non-zero $n$-th residues modulo $p$. Fix also positive integers $a_1,\ldots,a_k$ with sum $2k$. Then $$ \sum x_1^{a_1}\cdots x_k^{a_k} \equiv 2(-1)^{k-1}k! \bmod{p}, $$ where the sum is taken over all pairwise distinct $n$-residues $x_1,\ldots,x_k$ (counting the permutations).