Can I say that in an Artinian Ring $A$, if $\frak{m}$ is maximal, then $\mathfrak{m}^k$ is maximal $\forall k \in \mathbb{N}$?
This question arise because I would like to conclude that $\mathfrak{m_1}^k \cdots\mathfrak{m_n}^k = (0) \Rightarrow A $ is artinian $\iff$ it is noetherian. But I know that this is valid only if the product is a finite product of maximal ideals.
My attempt of a proof is: let's suppose that $\mathfrak{m}^k$ is not maximal, then $\exists \mathfrak{a} \subset A$ proper ideal such that $\mathfrak{m}^k \subset \mathfrak{a}$. So given that every maximal ideal is prime $\sqrt{\mathfrak{m}^k} = \mathfrak{m} \subset \sqrt{\mathfrak{a}} $ which is proper. Contradiction. Is it right?.