In Bob Vaughan's book The Hardy-Littlewood Method, early on he gives a sum
\begin{equation} \left(\sum_{m=1} ^N e(\alpha m^k)\right)^s = \sum_{m_1 = 1} ^N \sum_{m_2 = 1} ^N \cdots \sum_{m_s = 1} ^N e\left(\alpha(m_1 ^k + m_2 ^k + \cdots + m_s ^k)\right) \end{equation} where $N = [n^{1/k}]$, $\alpha$ is some real parameter and $k, s$ are integers. Then he says this equals \begin{equation} \sum_{m=1} ^{sn} R_s(m, n) e(\alpha m) \end{equation} where $R_s(m, n)$ is the number of representations of $m$ as the sum of $s$ $k$th powers, none of which exceed $m$.
I get where the $R_s$ term comes from but I don't see how the summation up to $N$ turns into $sn$ rather than $sN$, especially with the fact that, in general, $N^k \neq n$. Previously $n$ is taken to be sufficiently large but that only exacerbates the problem, as far as I can see.
Right, I see why. The additional terms between $sN^k$ and $n$ will be $0$ because of the condition that the $k$th powers have to be less than $n$ or, in other words, their $kth$ roots of the exponents are at most $[n^{1/k}]$