= One minus a nilpotent element is a unit
{title2=$(1-z)^{-1}=1+z+\cdots+z^{m-1}$}
If $z$ is a <nilpotent element> of a <ring>, choose $m\geq1$ with $z^m=0$. The finite <geometric series> identity gives
$$
(1-z)(1+z+\cdots+z^{m-1})=1-z^m=1,
$$
and the factors commute, so $1-z$ is a <unit>. In a <commutative ring>, every multiple of a <nilpotent element> is again <nilpotent>. Consequently an equality $x=ax^2$ with $x$ <nilpotent> forces $x=0$: rewrite it as $(1-ax)x=0$ and multiply by the inverse of $1-ax$.
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