@Shivering Soldier has already posted a link to a very nice direct proof of this fact using primary decomposition, and @RithvikReddy has already written a proof using the Artin-Rees lemma (+1), but here's another alternative proof.
Let $R$ be any (commutative) Noetherian ring and let $I$ be an ideal of $R$. We will show that $IJ=J$, where $J=\bigcap_{n=1}^\infty I^n$.
To see this, note that the set of ideals of $R$ whose intersection with $J$ is $IJ$ is non-empty, as it contains $IJ$. So, since $R$ is Noetherian, we may find an ideal $K\leqslant R$ maximal subject to $K\cap J=IJ$.
Claim: For any $x\in I$, there exists $k\in\mathbb{N}$ such that $x^k\in K$. To see this, consider the following ascending chain of ideals $$(K:x)\subseteq(K:x^2)\subseteq(K:x^3)\subseteq\dots.$$ Since $R$ is Noetherian, this chain must stabilize; let $k$ be such that $(K:x^l)=(K:x^k)$ for all $l\geqslant k$. Now we claim $x^k\in K$; otherwise, $K':=K+(x^k)$ is an ideal strictly containing $K$, and so by maximality of $K$ we have $K'\cap J\supsetneq IJ$, and so there exists $a\in K$ and $\lambda\in R$ such that $a+\lambda x^k\in J\setminus IJ$. In particular, note that $$xa+\lambda x^{k+1}=x(a+\lambda x^k)\in IJ\subseteq K,$$ so in fact, since $a\in K$, also $\lambda x^{k+1}\in K$, ie $\lambda\in (K:x^{k+1})$. But $(K:x^{k+1})=(K:x^k)$ by hypothesis, so in fact $\lambda x^k\in K$, whence $a+\lambda x^k\in K\cap J=IJ$, a contradiction. Thus indeed $x^k\in I$ and so we are done. $\blacksquare$
Now, since $R$ is Noetherian, $I$ is finitely generated, and so, by applying the claim to each of the generators of $I$ and taking the sum of the exponents, we can find $n\in\mathbb{N}$ such that $I^n\subseteq K$. Thus $I^n\cap J\subseteq K\cap J=IJ$. On the other hand, $J\subseteq I^n$ and so $I^n\cap J=J$, so we in fact have $J\subseteq IJ$ and thus we are done.
