I am trying to understand the proof for Artinian $\implies$ Noetherian given in chapter 16 of "Abstract Algebra (3rd edition)" by Dummit and Foote.
The proof given in the textbook goes as follows.
It suffices by (4) to prove that an Artinian local ring is Noetherian, so assume $R$ is Artinian with unique maximal ideal $M$. In this case we have $M = Jac R$, so $M^m= (JacR)^m$ for some m. Then $R\simeq R/M^m$ and it is an exercise to see that $R/M^m$ is Noetherian (cf. Exercise 8)
(4) Here is the previous proposition which states that every artinian is isomorphic to a finite direct product of local artinian rings. The proof for this gives the isomorphism $$ A\simeq A/M_1^m\times A/M_2^m\times ... \times A/M_n^m $$ where $M_1,..,M_n$ are the maximal ideals of $A$ and each $A/M_i^m$ is an artinian local ring with maximal ideal $M/M_i^m$.
And exercise (8) is: Let $M$ be a maximal ideal of the ring $A$ and suppose $M^n=0$ for some $n\geq1$. Then $A$ is Noetherian if and only it is Artinian.
I understand the proof for (4) and I have proven (8) but this proof still doesn't make to me.
We need to show that $A/M_i^n$, a local artinian ring, is noetherian. This concludes our proof because direct product of noetherian rings is noetherian. But how is $A/M_i^n$ proven to be noetherian?