Is this a valid proof of Nakayama's Lemma? I ask because I find the inductive step much more intuitive than the one I read in most authors...
Proposition (Nakayama): Suppose that $R$ is a commutative ring with unity, $I \subset R$ is an ideal contained in every maximal ideal, and $M$ is a finitely generated $R$ module. Suppose that $IM=M$. Then $M=0$.
Proof: Induction on the number of generators. If $M$ is cyclic and $M=Rm$, then there exists $i$ such that $im = m$, and then if $m\neq 0$ then $(1-i)m=0$ and $1-i$ is not a unit, contradicting that $i$ is contained in every maximal ideal (proof slightly abbreviated).
Inductive step: If the proposition is true for modules with $s$ generators, suppose $M$ is generated by $x_1, \dotsc, x_{s+1}$, and suppose this is a minimal generating set. Then let $\tilde{M}:=M/<x_{s+1}>$ is generated by $s$ elements and $I\tilde{M}=\tilde{M}$, so $\tilde{M}=0$, contrary to minimality of generators.