$1+a$ and $1-a$ in a ring are invertible if $a$ is nilpotent [duplicate]
Let $(A, +, \cdot)$ be a ring with $1$. An element $a\in A$ is nilpotent if there exists $n\in \mathbb{N}$ so that $a^n=0$.
Show that if $a$ is nilpotent then $1+a$ and $1-a$ are invertible.
Solution 1:
$1 = 1 - {a^n} = (1 - a)(1 + a + \cdots + {a^{n - 1}})$
If $n$ is an odd number it's the same for $1+a$: $1 = 1 + {a^n} = (1 + a)\sum\limits_{k = 0}^{n - 1} {{{( - 1)}^k}{a^k}} $.