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?

A slight variation on caffeinmachine's answer that I prefer, because I think it shows more of the structure of what's going on:
Let $F$ be a finite field (and thus has characteristic $p$, a prime).


*

*Every element of $F$ has order $p$ in the additive group $(F,+)$. So $(F,+)$ is a $p$-group.

*A group is a $p$-group iff it has order $p^n$ for some positive integer $n$.


The first claim is immediate, by the distributive property of the field. Let $x \in F, \ x \neq 0_F$. We have 
\begin{align}
p \cdot x &= p \cdot (1_{F} x) = (p \cdot 1_{F}) \ x
\\
& = 0
\end{align}
This is the smallest positive integer for which this occurs, by the definition of the characteristic of a field. So $x$ has order $p$.
The part that we need of the second claim is a well-known corollary of Cauchy's theorem (the reverse direction is just an application of Lagrange's theorem).
