Nilpotency order of a maximal ideal $M$ of a finite local ring $(R,M)$

Question

I need an argument for the following assertion.

If $(R, M)$ is a local ring of order $p^k$ for some prime integer $p$, then $M^{k}=0$. In particular, nilpotency order of $M$ is $<$ the order of $R$.

Answer 1

First of all, we may assume that $R\neq 0$ and $R$ is not a field. Consider the chain $M^k\subseteq M^{k-1}\subseteq \ldots \subseteq M^2\subseteq M$. If $M^i=M^j$ for some $1\leq i<j\leq k$, then $M^i=0$ and hence in particular, all greater powers of $M$ are zero, by Nakayama lemma. So, assume that the above chain is strict, that is, $M^k\subset M^{k-1}\subset \ldots \subset M^2\subset M$. Therefore order of all $M^i$'s are disncit $k$ numbers $\leq p^{k-1}$ and all of them divides $p^k$. Thus, order of $M^k$ must be equal to $1$, as desired.

Note that, if the order of $R$ is $p^k$, then the nilpotency order of $M$ is $\leq k$.