The uniqueness of a special maximal ideal factorization

The uniqueness of a special maximal ideal factorization

The following problem is from Michael Artin's Algebra, chapter 12, M.6,
unstarred:
Let $R$ be a domain, and let $I$ be an ideal that is a product of distinct
maximal ideals in two ways, say $I=P_1\dotsb P_r=Q_1\dotsb Q_s$. Prove
that the two factorizations are the same, except for the ordering of the
terms.
Well, following the proof of the fundamental theorem of arithmetic, it
seems that we should prove the theorem in the following way:
Show that $P_1=Q_j$ for some $j$.
Cancel $P_1,Q_j$ from both sides and reduce it to the induction hypothesis.
The first bulletin is relatively easy. It follows from the fact that
maximal ideals are prime, therefore if $M$ is maximal, and $M\supset AB$,
then $M\supset A$ or $M\supset B$, thus there's some $Q_j\subset P_1$,
therefore $Q_j=P_1$ following from the maximality of $Q_j$.
However, the second one seems hard. There seems no cancellation laws among
ideals. There's a counterexample even for maximal ideals, since maximal
ideals could be zero. It's not like prime number, which are always
positive. I should note that the distinctness of $P_k$,$Q_j$ and the fact
that $R$ is a domain isn't used.
So how can we proceed next? Any help? Thanks!