Here's my attempt at it:
Let $A$ be a finite integral domain and $a\in A$ and suppose that $A=\langle a \rangle$. Hence, $|a|=|A|$. By definition of the order of an element, we have that $|a| \cdot a = |A| \cdot a = 0$. If $p$ is the characteristic of $A$ (which must be prime), then $p$ divides $|A|$. Also since $p$ is the characteristic of $A$ it must be that $p \cdot a = 0$. Hence $|A| = |a|$ divides $p$. Thus, we have $p=|A|$.
Is this proof okay? This question is from Charles Pinter's A book of Abstract Algebra.
yep it is!
I would have gone over the intermediate step of proving it is a field. And since I feel bad just putting "yep it is" as answer, I just attach it:
as $A$ is integral domain, $x\cdot: A \to A$ is injective for all $x\neq 0$, in particular since $A$ is finite, it is invertible. But that means that $R$ is a field. So it is a finite field, and hence has a primepower of elements. Now since it is cyclic this means that it is isomorphis to $\mathbb{Z}/p^n\mathbb{Z}$. but for this to be a field we need $n=1$ and we are done.
Respectively, if you are unhappy with the argument about $\mathbb{Z}/p^n\mathbb{Z}$, one can also (since one now knows precisely the structure of $A$) argue by saying that if it is not a prime number it is a vector space over a prime field and hence can't be cyclic.