Let $M$ be an $R$-module of finite length and $N$ a maximal submodule of $M.$ 



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It is a generalization of this result:


In a Notherian ring $R,$ all minimal prime ideals have non-zero annihilator.



Annihilator of minimal prime ideal in a commutative Noetherian ring
Put $P=(N:M)$. 


Because $N$ is a maximal submodule, $P$ is a maximal ideal. 


Because $M$ has finite length, there is some minimal $k$ such that $MP^k=MP^{k+1}$. By Nakayama, there exists $s\in 1+P$ such that $MsP^k=0$. 


Put $T=MsP^{k-1}$. If $T=0$, then $MP^{k-1}\subset MP^k$, contradicting minimality of $k$. So $T\neq 0$. But by the preceding paragraph, $TP=0$. 


So if $k\neq 0$, then any nonzero $m\in T$ will do. But if $k=0$, then $Ms=0$. Therefore $s\in ann(M)\subseteq (N:M)=P$ and hence $1\in P$, a contradiction.


