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Let $R$ be a ring with unity, and suppose $x\in R$ is nilpotent $(i.e. x^n=0$ for some positive integer $n$ $)$. Prove that $1-x$ is a unit in $R$.

Any hints or proofs are greatly appreciated. Rings are really new to me.

Yuri
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Multiply it with $$ 1+x+x^2+\cdots +x^{n-1} $$ You end up with $1-x^n$, which equals $1$, since $x^n=0$.

Since we have now proven that there is a $y$ such that $(1-x)y=1$, by definition of "unit", $1-x$ is a unit.

Arthur
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    I just thought I would add in something about the motivation for the inverse: $(1-x)^{-1}$ can be thought of as $\frac{1}{1-x}$ is like the geometric series with ratio $x$. This is where you get $1+x+x^2 + x^3 + \ldots$. Nilpotent makes this "infinite" series terminate after finitely many terms so you don't need to worry about convergence. Otherwise this inverse might seem somewhat magical. – CPM Oct 26 '14 at 23:39