I'm trying to prove the following particular case of Nakayama's lemma. Let $R$ be a commutative ring and $a\in R$ be nilpotent (let's suppose $a^{k-1}\not=0$, $a^k=0$). Then $aM=M \Rightarrow M=\{0\}$.
I have done this. Since $R$ is commutative, $aM\subset M$ is a submodule.
Let $m\in M$. Then $m=an$ for some $n\in M$. If I act by $a^{k-1}$ I get: $a^{k-1}m=a^kn=0$
So I get that $0\ne a^{k-1} \in Ann(M)$, where $Ann$ denotes the annihilator. I don't know where to go from here.