$|S(p,a)|\le (k-1)\sqrt{p}$ for prime moduli

Let $p$ be a prime, $k\ge1$, and $a$ an integer with $p\nmid a$. Then the complete exponential sum over a prime modulus satisfies the square-root cancellation bound

Context. Write $d=\gcd(k,p-1)$. The $k$-th powers modulo $p$ coincide with the $d$-th powers, and for $y\not\equiv0$ the number of solutions of $x^{k}\equiv y$ is $\sum_{\chi^{d}=\chi_{0}}\chi(y)$, the sum running over the $d$ multiplicative characters of order dividing $d$. Substituting and separating the principal character (whose contribution cancels the $m\equiv0$ term) expresses $S(p,a)$ as a sum of $d-1$ Gauss sums,

each of modulus exactly $\sqrt{p}$. Since $d\le k$, the bound follows.

Role. This is the one place in the proof of Vaughan's Theorem 4.2 where the trivial bound is not enough: for $h=1$ the estimate $|S(p^{h},a)|\le p^{h}$ gives $p^{1/k}$ too much, and no elementary substitution recovers it. Every higher prime power reduces to this case by the recursion $S(p^{h},a)=p^{k-1}S(p^{h-k},a)$. Note that $\tfrac12\le 1-\tfrac1k$ for $k\ge2$, so square-root cancellation is comfortably stronger than the $p^{1-1/k}$ ultimately needed.

Degenerate cases. When $\gcd(k,p-1)=1$ the map $x\mapsto x^{k}$ permutes the residues, so $S(p,a)=0$; the stated bound is then an equality only if $k=1$. The right-hand side is $\ge0$ because $k\ge1$.