Proving the spectrum of a commutative Artinian ring is finite This proof was done in my class of Ring Theory and I have 1 question in 1 part of the proof.

Theorem: Let A be an artinian (commutative ) ring. Then show that: (1)  The jacobson radical $M_A$ of A is nilpotent. (2) Spm A is a finite set. (3) A is a noetherian ring.

I have question only in (2) of the proof. It has been proved that distinct maximal ideals are comaximal.  Let $M_1 ,..., M_s \in Spm A$ be distinct maximal ideals and consider $M_1M_2...M_s \subseteq  ...\subseteq M_1 M_2\subseteq M_1\subseteq A$. I understand how this chain is descending   but I am not able to prove that this chain is strictly descending  at each place.
Can you please help me with the proof?
 A: The condition that $A$ is Artinian means that if you have a sequence
$$
I_0\supseteq I_1\supseteq I_2\supseteq \dots \supseteq I_n\supseteq \dotsb
$$
of ideals, then there exists $m$ such that, for all $n\ge m$, it holds that $I_n=I_m$.
An equivalent formulation is that no infinite strictly decreasing sequence of ideals exists, but the above is generally easier to use. However, for this case, another characterization is more suited, namely

the ring $A$ is Artinian if and only if every non empty set of ideals has a minimal element.

Take the set $\mathcal{M}$ consisting of ideals that are (finite) products of maximal ideals. Since a maximal ideal exists, $\mathcal{M}$ is not empty. Take $M_1M_2\dots M_r$ a minimal element and any maximal ideal $M$. Since
$$
MM_1M_2\dots M_r\subseteq M_1M_2\dots M_r
$$
by minimality we deduce equality and therefore that
$$
M_1M_2\dots M_r\subseteq M
$$
Since $M$ is maximal, it is prime and so it holds $M_i\subseteq M$ for some $i$. Maximality of both implies $M=M_i$ and so $\{M_1,M_2,\dots,M_r\}$ is the whole set of maximal ideals.
