Let $R$ be a commutative ring with $1$. Let $a\in R^\times$ and $b\in\text{Nil}(R)$. Show that $a+b\in R^\times$.
Attempts:
Suppose $b^n=0$. I've tried to figure out a "simple" case as $n=2$;
I've tried to develop the expressions $(a+b)^2=a(a+2b)$ and $a^{-1}(a+b)=1+a^{-1}b$ but nothing came up.
Also $\exists x,y\in R$ s.t. $1=xa+yb$ but it doesn't help because we get $x=a^{-1},y=0$.
Another direction i've failed is to show $\langle 0\rangle=R/\langle a+b\rangle$.
Hint the inverse of $1-r$ is $1+...+r^{n-1}$ if $r^n=0$.
If $a$ is invertible, $a+r=a(1+a^{-1}r)$ and $a^{-1}r$ is nilpotent. The product of invertibles elements is invertible.