Nakayama's Lemma/Proof 2

Proof
Let $\phi : M \to M$ be the identity mapping on $M$, i.e.:
 * $\forall x \in M : \map \phi x = x$

Since $\mathfrak a M = M$, $\phi$ is an endomorphism of $M$ such that:
 * $\map \phi M \subseteq a M$

By Cayley-Hamilton Theorem, there exist $a_0, \ldots, a_{n-1} \in \mathfrak a$ such that:
 * $(1):\quad \phi^n + a_{n - 1} \phi^{n-1} + \cdots + a_1 \phi + a_0 = 0$

Evaluating $(1)$ at $x=1$, especially, we have:
 * $(2):\quad 1 + a= 0$

where:
 * $a := a_{n - 1} + \cdots + a_1 + a_0$

Since:
 * $a \in \mathfrak a \subseteq \map {\operatorname {Jac} } A$

by Characterisation of Jacobson Radical, $1 + a$ is a unit in $A$.

Thus for each $x \in M$, we have:

That is:
 * $M = 0$