TITLE: Order of finite fields is $p^n$
QUESTION [63 upvotes]: 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?

REPLY [4 votes]: Let $n,m$ be positive integers and $F$ be a finite field. Define the operations.
\begin{align}
(n \cdot \mathbb{1}_F) (m \cdot \mathbb{1}_F) &= (nm \cdot \mathbb{1_F})\\
(-n)\cdot \mathbb{1}_F &= -(n\cdot \mathbb{1}_F)\\
0 \cdot \mathbb{1}_F &= 0\\
\end{align}
This then suggests there is a natural homomorphism 
\begin{align}
\mathbb{Z} &\stackrel{\phi}\to F\\
n &\to n \cdot \mathbb{1}_F
\end{align}
Now since $F$ is finite, it has prime characteristic $p$ and we may see that $\ker \phi = p\mathbb{Z}$. Subsequently, we also see 
$$\mathbb{Z}/\ker\phi = \mathbb{Z}/p\mathbb{Z} \stackrel{\phi}\to \text{img} \phi \approx F_p \subset F$$
and in fact $F$ is a finite extension of $F_p$ (say $[F: F_p] = n$), that is, $F_p \leq F$ or in other words, $F$ is a vector space over $F_p$. So any element $z \in F$ can be written as a linear combination of basis elements $(z_1, \dots, z_n) \subset F$ and $a_i \in F_p$
$$z = a_1z_1 + \dots a_nz_n$$
Now by counting, each $a_i$ has $p$ choices. So going through $n$ basis elements, there are exactly $p^n$ total choices and this counts all the $z$s in $F$. Therefore $|F| = p^n$