Can you check my Work or mention a link for a proof.
Let $R$ be Noetherian ring. Then if $M$ is a maximal ideal in $R$. Prove that $M^2 \ne M$.
Proof:
Since $R$ is Noetherian ring then $M$ is finitely generated. Thus $M = (a_1, a_2 , \cdots, a_k)$ we can choose the $a_i's $ which are minimal...