If $a \in A$ maps to a unit of $A/\operatorname{nil}(A)$, then $a$ is a unit of $A$

Question

Let $A$ be a ring, $I \subset \operatorname{nil}(A)$ an ideal made up of nilpotent elements. Show that if $a \in A$ maps to a unit of $A/I$, then $a$ is a unit of $A$.

I tried this: If $ab = 1 - x$ with $x^n=0$. How can I prove that $$ ab(1+x+\ldots+x^{n-1}) = 1 \, ? $$

Hope someone could help me please. Thanks for your time and help.

Answer 1

Hint: \begin{align*} (1-x)(1 + x + x^2 + \cdots + x^{n-1}) = 1 &+ x + x^2 + \cdots + x^{n-1}\\ &- x - x^2 - \cdots - x^{n-1} - x^n = \cdots \end{align*}