= Solution
Let $I=(a_1,\ldots,a_s)$ and form the <Rees algebra> $\mathcal R=\bigoplus_{j\ge0}I^jt^j$. It is generated over $R$ by $a_1t,\ldots,a_st$, so the <Hilbert basis theorem> makes it <Noetherian>. The <Rees module> $\mathcal M=\bigoplus_{j\ge0}I^jMt^j$ is finite over it, generated by finitely many generators of the <module> $M$ in degree zero. Hence its graded <submodule>
$$
\mathcal N=\bigoplus_{j\ge0}(N\cap I^jM)t^j
$$
has finitely many homogeneous generators, all in degrees at most some $k$.
For $i\ge k$, each degree-$d$ generator contributes a multiple from $I^{i-d}(N\cap I^dM)$. This is contained in $I^{i-k}(N\cap I^kM)$, because multiplication by $I^{k-d}$ puts its coefficient both in $N$ and in $I^kM$. Thus $N\cap I^iM\subseteq I^{i-k}(N\cap I^kM)$. The reverse inclusion follows directly by multiplication: it stays in $N$ and in $I^iM$. Consequently
$$
\boxed{N\cap I^iM=I^{i-k}(N\cap I^kM)\qquad(i\ge k)}.
$$
This is the <Artin-Rees lemma>, proved through finite graded generation rather than an assumption that the intersection filtration initially has that form.
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