If $r$ is a nilpotent element of a ring $R$ prove or disprove that $1-r^n\in U(R)$

Question

I'm not sure if the end of induction is correct (need some feedback on that spot)> If $r$ is a nilpotent element of a ring $R$ prove or disprove that $1-r^n\in U(R)$

I'm trying to prove the positive

My solution (Symbolism) $U(R)$ the set of all invertible elements in $R$

The element $r\in R$ has the property: There is a minimum number $k\in\mathbb{N}$ such that $r^k=0_R$ (1) $1_R=1_R-r^k$. If $k\leq{n}$ the statement stands since $r^k=0_R$. We want to check the case where $n<k$.

We will work with induction.

For $n=1$ we have $1-r^1=1-r$. Using (1) we proceed as $1_R=1_R-r^k=(1-r)(1+r+r^2+...r^{k-1})$ so $1-r\in U(R)$. We assume that $1-r^n\in R$. We will prove that $1-r^{n+1}\in U(R)$.Indeed, we have $1-r^{n+1}=(1-r)(1+r+r^2+...+r^n)$ where the second factor sums to $\frac{1}{1-r}$ as a geometric partial sum. Thus, $1-r^{n+1}=\frac{1-r}{1-r}=1_R\in U(R)$. Then for for every $n\in\mathbb{N}$: $1-r^n\in U(R)$ with the condition that $r\in R$ is nilpotent. Is it correct?