If $\text{Ann}_R(r)=\text{Ann}_R(r^n)$ for each $r\in R$ and for each $n\in \Bbb N$, then show that $r^n=0$ implies $r=0$ for each $r\in R$.

Question

Let $R$ be a ring with unit (not necessarily commutative) and $\text{Ann}_R(r)=\{r'\in R|r'r=0\}$ . If $\text{Ann}_R(r)=\text{Ann}_R(r^n)$ for each $r\in R$ and for each $n\in \Bbb N$, then show that $r^n=0$ implies $r=0$ for each $r\in R$.

Clearly $a^n=0$ implies $a\in \text{Ann}_R(r^{n-1})=\text{Ann}_R(r)$ so that $r^2=0$. I cannot proceed from there.

Answer 1

Simply if $a^n=0$ then the identity element $1\in{\rm Ann}_R(a^n)$ and this is equal to ${\rm Ann}_R(a)$, so $a=a\cdot 1=0$.