If $x=0$, then $1\!+\!x=1$, which is a unit. Thus, we may assume that $x\neq 0$.

Since $x$ is nilpotent, there is a positive integer $n$ such that $x^{n}=0$. We multiply $1\!+\!x$ by another ring element:

(Note that the summations include the term  $(-1)^{0}x^{0}$, which is why $x=0$ is excluded from this case.)

The reversed multiplication gives the same result. Therefore, $1\!+\!x$ has a multiplicative inverse and thus is a unit. ∎