Showing that this ring's order is a power of $2$

Question

Let $(A, +, \cdot) $ be a finite unital ring. If $2$ is nilpotent, then $A$'s order is a power of $2$.
The answer key suggests that this implies that the order of any element in the additive group is of the form $2^m$ where $m \in \mathbb{N}$, but I don't see why this is the case. I tried using Cauchy' s theorem, but to no avail.

Answer 1

Since $2$ is nilpotent, $2^n=0$ for some $n\in\mathbb{N}$. Now fix $x\in A$ and let $d$ be the additive order of $x$. Since $2^nx=0x=0$, $d$ divides $2^n$, so it is a power of $2$.