If $F$ is a finite field then $|F|=p^n$ for prime $p$ and some integer $n$

Question

I need some help proving the following:

If $F$ is a finite field then $|F|=p^n$ for prime $p$ and some integer $n$.

By contradiction, suppose $|F| =pq$ for primes $p$ and $q$. Then by Cauchy's theorem there exists $a, b\in F$ such that the order of $a$ is $p$ and the order of $b$ is $q$. I am not quite sure where to go from here, but I think we need to contradict that $F$ is a finite field. The only thing that comes to mind is to show zero-divisors.

1. Where can I go from here?
2. I thought every finite field had to have prime order? $p^n$ is not prime!

Thanks for all the help!

Answer 1

Think of $F$ as a vector space over its smallest subfield containing $1$ (which must be ...?).