Order of finite fields is $p^n$ Let $F$ be a finite field.
How do I prove that the order of $F$ is always of order $p^n$ where $p$ is prime?

Let $F$ be a finite field. Then the underlying additive group of the field (let's denote this by $F^+$) has this interesting property:

For every two non-identity (i.e. non-zero) elements $a$ and $b\in F^+$, there is an automorphism $\phi$ of the additive group such that $\phi(a)=b$.

This can be seen by examining the map $(x\mapsto ba^{-1}x)$.
This means the set of automorphisms of $F^+$ act transitively on $F^+$. Since automorphisms permute elements of the same order, we can conclude that every element in $F$ has the same order.
But a finite group in which all non-identity elements have the same order is necessarily a $p$-group such that every element has prime order. This can be shown by Cauchy's Theorem.
Suppose the order of $F^+$ had two distinct prime factors $p$ and $q$. Then $F^+$ would contain an element of order $p$ and another element of order $q$ by Cauchy's Theorem. This contradicts that every element has the same order. So the order of $F$ is indeed a prime-power. Cauchy's Theorem implies that $F$ has an element of order $p$, thus all elements have order $p$ by the hypothesis.
So, $F$ must be of prime-power order $p^n$, and we have that $px=0$ for all $x\in F$.
