$|S(p^h,a)|\le k^{3}\,(p^h)^{1-1/k}$ at prime powers

Let $k\ge2$, let $p$ be a prime, let $h\ge1$, and let $a$ be an integer with $p\nmid a$. Then

This is the local half of Vaughan's Theorem 4.2: the estimate $|S(q,a)|\ll q^{1-1/k+\varepsilon}$ restricted to prime-power moduli, with an explicit constant depending only on $k$.

The proof is an induction on $h$ in which three regimes appear. For $h=1$ one needs genuine cancellation, supplied by the Gauss-sum bound $|S(p,a)|\le(k-1)\sqrt p$ together with $\tfrac12\le1-\tfrac1k$. For $h$ below $2(v_{p}(k)+1)$ the trivial bound $|S|\le p^{h}$ already suffices, because $p^{v_p(k)}\mid k$ forces $p$ and $h$ to be bounded in terms of $k$ alone — this is where the constant $k^{3}$ comes from. Above that threshold the recursion $S(p^{h},a)=p^{k-1}S(p^{h-k},a)$ applies, and it is exponent-neutral: $p^{k-1}\cdot p^{(h-k)(1-1/k)}=p^{h(1-1/k)}$, so the induction closes with the same constant. When $h\le k$ the recursion instead terminates at the exact value $p^{h-1}$, which already satisfies the bound.

Combined with multiplicativity of $q\mapsto S(q,a)$ and the divisor bound, this yields the global estimate for arbitrary moduli.