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 Sequence in Ring:
 * $\paren {1 + a} \cdot \ds \sum_{k \mathop = 0}^{n - 1} \paren {-a}^k = 1 + \paren {-a}^n$
 * $\paren {\ds \sum_{k \mathop = 0}^{n - 1} \paren {-a}^k} \cdot \paren {1 + a} = 1 + \paren {-a}^n$

where $\sum$ denotes summation.

By Negative of Nilpotent Ring Element:
 * $\paren {-a}^n = 0$

Thus $1 + a$ is a unit.

Also see

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