proving that $char(A) =p$ for some prime number $p$ Let A be an integral domain. If the order of $A$ is a prime number $p$, then the characteristic of $A$ must be equal to $p$.
Here's what I got so far. BY definition, for $a \in A, a \neq 0$, we know that the order of $a$ is the least positive integer such that $a^p =1$ (Since $A$ is a ring with unity)
$a^p = a \cdot a \cdot a \cdot ..... \cdot a = 1$ but I am stuck as to where to go from here. 

Let $m$ be the order of $1$ in the additive group of $A$. By LaGrange's Theorem, $m$ must divide $\vert A \vert = p$. Therefore $m = 1$ or $m = p$. Since $A$ is not the zero ring, we must have $m = p$, hence $\textrm{char}(A) = p$.
It's worth noting that this also implies that $A$ is generated by $1$ as an additive group, hence $A \cong \mathbb{Z}/p\mathbb{Z}$.
