TITLE: Prove $1+x$ is a unit in $R: \text{commutative ring}$ where $x$ is nilpotent
QUESTION [6 upvotes]: Prove $1+x$ is a unit in $R: \text{commutative ring}$ where $x$ is nilpotent
do I need to make use of a Taylor series expansion for this?
$(1+x)(1+x)^{-1} = 1 \implies (1+x)^{-1} = \displaystyle\frac{1}{1+x}$
and $x$ being nilpotent then for some positive integer n, $x^n=0$ I need to make use of this. 
Or is there another way to do it?

REPLY [8 votes]: If $x^n=0$, then $$(1+x)\left(\sum_{k=0}^{n-1} (-1)^kx^k\right)=1+(-1)^{n-1}x^n=1.$$

REPLY [6 votes]: Hint:
$$(1+x)(1-x+x^2-\cdots\pm x^{n-1})=1\pm x^n=\cdots?$$

REPLY [3 votes]: You are exactly right about using the geometric series expansion. The point is the terms with high enough power are zero since $x$ is nilpotent, so it is actually a finite sum. Suppose $x^n=0$. Then $(1+x)(1-x+x^2+...+(-1)^{n-1} x^{n-1})=1$