For $n=1$ this is clear, since $M$ is an ideal of $R$ and is generated by some element $a\in R$. Now suppose that the statement is true for all submodules of $R^{m},1\leq m\leq n-1$.

For a submodule $M$ of $R^{n}$ we define $f\colon M\to R$ by $(k_{1},\ldots,k_{n})\mapsto k_{1}$. The image of $f$ is an ideal $\mathfrak{I}$ in $R$. If $\mathfrak{I}=\{0\}$, then $M\subseteq\ker(f)=(0)\times R^{{n-1}}$. Otherwise, $\mathfrak{I}=(g),g\neq 0$. In the first case, elements of $\ker(f)$ can be bijectively mapped to $R^{{n-1}}$ by the function $\ker(f)\to R^{{n-1}}$ given by $(0,k_{1},\ldots,k_{{n-1}})\mapsto(k_{1},\ldots,k_{{n-1}})$; so the image of $M$ under this mapping is a submodule of $R^{{n-1}}$, which by the induction hypothesis is finitely generated and free.

Now let $x\in M$ such that $f(x)=gh$ and $y\in M$ with $f(y)=g$. Then $f(x-hy)=f(x)-f(hy)=0$, which is equivalent to $x-hy\in\ker(f)\cap R^{n}:=N$ which is isomorphic to a submodule of $R^{{n-1}}$. This shows that $Rx+N=M$.

Let $\{g_{1},\ldots,g_{s}\}$ be a basis of $N$. By assumption, $s\leq n-1$. We’ll show that $\{x,g_{1},\ldots,g_{s}\}$ is linearly independent. So let $rx+\sum_{{i=1}}^{s}r_{i}g_{i}=0$. The first component of the $g_{i}$ are 0, so the first component of $rx$ must also be 0. Since $f(x)$ is a multiple of $g\neq 0$ and $0=r\cdot f(x)$, then $r=0$. Since $\{g_{1},\ldots,g_{s}\}$ are linearly independent, $\{x,g_{1},\ldots,g_{s}\}$ is a generating set of $M$ with $s+1\leq n$ elements. ∎

Let $\{g_{1},\ldots,g_{s}\}$ be a generating set of $M$ and $f\colon R^{s}\to M$, $(r_{1},\ldots,r_{s})\mapsto\sum_{{i=1}}^{s}r_{i}g_{i}$. Then the inverse image $N^{{{}^{{\prime}}}}$ of $N$ is a submodule of $R^{s}$, and according to lemma finitely generated modules over a principal ideal domain can be generated by $s$ or fewer elements. Let $n_{1},\ldots,n_{t}$ be a generating set of $n^{{{}^{{\prime}}}}$; then $t\leq s$, and since $f$ is surjective, $f(n_{1}),\ldots,f(n_{t})$ is a generating set of $N$. ∎