This is the same as this question, though a solution was never given, and my approach is different than the hint.
It's also the first exercise in Commutative Ring Theory by Matsumura.
Let $A$ be a (commutative) ring and $I \subset \operatorname{nil}(A)$ an ideal made up of nilpotent elements; if $a\in A$ maps to a unit of $A/I$ then $a$ is a unit of $A$.
Attempt:
Since $a$ maps to a unit in $A/I$, we have that for $a+I$ there exists some coset $a'+I$ such that $(a+I)(a'+I)=1+I$. By coset multiplication, we have $(aa')+I=1+I$, which implies that $aa'= 1+i$ for some $i\in I$. Since $i$ is nilpotent, there exists $n>0$ such that $i^n=0$. If $i=0$, we are done. Otherwise, we have that $n \geq 2$ so we can do the following:
\begin{align*}
aa' &= 1+ i\\
aa'i^{n-1}&=(1+i)i^{n-1}\\
aa'i^{n-1} &= i^{n-1}+0\\
aa'i^{n-1} &= i^{n-1}\\
aa' &= 1,
\end{align*}
showing that $a$ is indeed a unit in $A$.
Is this correct?
P.S. I would guess that there is an easier method, but I wanted to try to proceed directly from the definition and work with cosets in the quotient ring.
Best Answer
\begin{align*} aa' &=1+i\\ aa'-1&=i\\ (aa'-1)^n&=i^n\\ (aa'-1)^n&=0\\ \sum_{k=0}^n {n\choose k}(aa')^k(-1)^{n-k}&=0\\ a\left(\sum_{k=1}^n {n\choose k}a^{k-1}a'^k(-1)^{n-k}\right)&=(-1)^{n+1}\\ a\left(\sum_{k=1}^n {n\choose k}a^{k-1}a'^k(-1)^{1-k}\right)&=1 \end{align*}