Nilpotent Ring Element plus Unity is Unit

Theorem
Let $A$ be a ring with unity.

Let $1 \in A$ be its unity.

Let $a \in A$ be nilpotent.

Then $1 + a$ is a unit of $A$.

Proof
Because $a$ is nilpotent, there exists a natural number $n > 0$ with $a^n = 0$.

By Sum of Geometric Progression in Ring:
 * $\left({1 + a}\right) \cdot \displaystyle \sum_{k \mathop = 0}^{n - 1} \left({-a}\right)^k = 1 + \left({-a}\right)^n$
 * $\left({\displaystyle \sum_{k \mathop = 0}^{n - 1} \left({-a}\right)^k}\right) \cdot \left({1 + a}\right) = 1 + \left({-a}\right)^n$

where $\sum$ denotes summation.

By Negative of Nilpotent Ring Element, $\left({-a}\right)^n = 0$.

Thus $1 + a$ is a unit.

Generalizations

 * Nilpotent Ring Element plus Unit is Unit
 * Topologically Nilpotent Element plus Unity is Invertible