Order of finite field is $p^{n}$, where $p$ is prime.

$F$ : finite field.
$char(F)$ : characteristic of $F$

We already know that $(F,\ +)$ is the group under the addition. Since that every field is integral domain,  characteristic of $F$ must be $0$ or some prime $p$.

Characteristic of finite field could not be $0$. Since that, by $\textit{Lagrange's Theorem}$, the order of each element of the group divides the order of the group if the order of the group is finite.

So, let's consider the case when $char(F)$ is some prime $p$. Then, $0 = 1 \cdot p$ which implies $|1| = p$ on $(F,\ +)$ and $p \mid |F|$.

Let's assume that there is some other prime $q$, where $q \mid |F|$. By the $\textit{Cauchy's Theorem}$, if $q \mid |F|$, then $|x| = q, \exists x \in F$.

If $char(F) = p$, then $x \cdot p = 0$ and $x \cdot q = 0$ must be satisfied. But, from the fact that $gcd(p,\ q) = 1$, by $Bezout's identity$, there exists some integers a, b which satisfies $ap + bq = 1$.

By multiplying $x$ on both sides of equation yields,

$(ap + bq) \cdot x = x$
$apx + bqx = x$
$axp + bxq = x$
 0 + 0 = x

Which is the contradiction to the $|x| = q$.

$$\therefore |F| = p^{n},\text{ for some prime }p,\ n > 0$$