Characteristic of Finite Ring with No Zero Divisors/Proof 2

Proof
Suppose $\operatorname{Char} \left({R}\right) = n$ where $n$ is composite.

Let $n = r s$, where $r, s \in \Z, r > 1, s > 1$.

First note that:

Then:

... as $R$ has no proper zero divisors.

But both $r$ and $s$ are less than $n$ which contradicting the minimality of $n$.

So if $\operatorname{Char} \left({R}\right) = n$ it follows that $n$ must be prime.

Now let $x \in R^*$.

Then by Characteristic times Ring Element is Ring Zero, $n \cdot x = 0_R$.

It follows from Element to Power of Multiple of Order is Identity that $\left|{x}\right| \backslash n$.

Since $n$ is prime, $\left|{x}\right| = 1$ or $\left|{x}\right| = n$.

It cannot be $1$, from Null Ring iff Characteristic is One, so the result follows.