The tensor product of two flat right modules is flat

Let $R$ and $S$ be rings with 1. Let $M$ be a right $R$-module and let $N$ be an $(R,S)$-bimodule. Prove that if $M$ is flat over $R$ and $N$ flat over $S$, then $M \otimes_R N$ is flat over $S$.

Let $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ be a short exact sequence of left $S$-modules. Since $N$ is flat over $S$, the natural sequence $0 \rightarrow N \otimes_S A \rightarrow N \otimes_S B \rightarrow N \otimes_S C \rightarrow 0$ is short exact. Moreover, this is a sequence of left $R$-modules since $N$ is an $(R,S)$-bimodule. Since $M$ is flat over $R$, $0 \rightarrow M \otimes_R (N \otimes_S A) \rightarrow M \otimes_R (N \otimes_S B) \rightarrow M \otimes_R (N \otimes_S C) \rightarrow 0$ is short exact. Using the natural isomorphisms, $0 \rightarrow (M \otimes_R N) \otimes_S A \rightarrow (M \otimes_R N) \otimes_S B \rightarrow (M \otimes_R N) \otimes_S B \rightarrow 0$ is short exact.

Thus $M \otimes_R N$ is flat as a right $S$-module.