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?

Suppose for contradiction that $$1-ab$$ is contained in a maximal ideal $$M$$. Since $$a$$ is in the Jacobson radical, $$a\in M$$, so $$ab\in M$$, and thus $$1 = (1-ab) + ab\in M$$, contradicting the fact that $$M$$ is proper.
Now we know that $$1-ab$$ is not contained in any maximal ideal. Let $$I = (1-ab)$$ be the principal ideal generated by $$1-ab$$. Since every proper ideal is contained in a maximal ideal, $$I$$ must be improper, so $$(1-ab) = A$$.
Suppose $$1-ab$$ is contained in maximal ideal $$I$$.
Then since $$a$$ is in each maximal ideal (and so is the multiple $$ab$$), $$(1-ab) + ab = 1$$ would be in maximal ideal $$I$$.