I have encountered the following proof that shows if an element $a$ of a ring $A$ is in the Jacobson radical $J(A)$, intersection of all maximal ideals, then $1-ab$ must be invertible $\forall b \in A$. Here is the proof (taken from "Commutative algebra" by Andrea Ferretti).
I don't quite get the first line of the proof:
"if $a \in J(A)$ then $1-ab$ is not contained in any maximal ideal, so $<1-ab> = A$".
What is the reasoning behind this?