How can I show that if $x$ is a nonunit then $1-x$ is a unit?

Question

Let $R$ be a commutative local ring. I want to show that then $x\in R$ is not a unit implies $1-x$ is a unit.

My idea was the following:

Since $R$ is a local ring we can chose $\mathfrak{m}$ to be it's unique maximal ideal. Now let $x\in R$ be a nonunit and assume that $1-x$ is also a nonunit. Then $I:=(1-x)\subsetneq R$. Since $\mathfrak{m}$ is a maximal ideal $I\subseteq \mathfrak{m}$. Now I wanted to write $1=1-x+x$ and I can deduce that $1-x+x$ is a unit. But somehow I don't see how to conclude from there. Can someone help me?

Answer 1

1. If $R$ is local with maximal ideal $\mathfrak{m}$, then $R^{\times} = R \setminus \mathfrak{m}$. See for example SE/3913530.
2. If $R$ is any ring and $I \subseteq R$ is any proper ideal, then $x \in I$ implies $1-x \notin I$. In fact, if $1-x \in I$, then $1 = x + (1-x) \in I$, so $I = R$.

The claim follows from 1. and 2.