Nontrivial ideal of a Noetherian domain contains a finite product of nonzero prime ideals

Consider the collection of nontrivial ideals of $R$ which do not contain a product of nonzero prime ideals. If this collection is nonempty, then it contains a maximal element $I$. I claim that $I$ must be prime (which is absurd).

Suppose otherwise, so that there exist $x,y \notin I$ such that $xy\in I$. The ideals $I+xR$, $I+yR$ are strictly larger than $I$, and they are not trivial: if, say, $I+xR=R$ then $yI+yxR = yR$ but $yI+yxR$ is contained in $I$, contradicting that $y \notin I$. Therefore by the choice of $I$ each of them must contain a product of prime ideals, but then so does $(I + yR)(I+yR) \subseteq I$, which contradicts the choice of $I$.

Theorem: Ideals of Noetherian rings have finite primary decompositions.

Lemma: If $Q$ is a primary ideal in a Noetherian ring whose radical is $P$ (which is prime, of course), then there exists a natural number n such that $P^n\subseteq Q$.

Hint: If $P_1,\dots P_m$ is a complete list of radicals of primary ideals $Q_1,\dots Q_m$ in the decomposition of your ideal $I$, then you can find an $N$ large enough so that $P_1^N\cdot\ldots \cdot P_m^N\subseteq \cap Q_i=I$.

Notice that this doesn't depend on $R$ being a domain or even reduced. (It does of course, depend on foreknowledge of primary decompositions and primary ideals, though.)