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 $p$ be the characteristic of a finite field $F$.${}^{\text{Note 1}}$ Then since $1$ has order $p$ in $(F,+)$, we know that $p$ divides $|F|$. Now let $q\neq p$ be any other prime dividing $|F|$. Then by Cauchy's Theorem, there is an element $x\in F$ whose order in $(F,+)$ is $q$.
Then $q\cdot x=0$. But we also have $p\cdot x=0$. Now since $p$ and $q$ are relatively prime, we can find integers $a$ and $b$ such that $ap+bq=1$.
Thus $(ap+bq)\cdot x=x$. But $(ap+bq)\cdot x=a\cdot(p\cdot x)+b\cdot(q\cdot x)=0$, giving $x=0$, which is not possible since $x$ has order at least $2$ in $(F,+)$.
So there is no prime other than $p$ which divides $|F|$.

Note 1: Every finite field has a characteristic $p\in\mathbb N$ since, by the pigeonhole principle, there must exist distinct $n_1&lt; n_2$ both in the set $\{1, 2, \dots, \lvert F\rvert +1\}$ such that $$\underbrace{1+1+\dots+1}_{n_1}=\underbrace{1+1+\dots+1}_{n_2},$$ so that $\underbrace{1+1+\dots+1}_{n_2-n_1}=0$. In fact, this argument also implies $p\le n$.
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 &amp;= p \cdot (1_{F} x) = (p \cdot 1_{F}) \ x
\\
&amp; = 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).