With any field, either $char(K) = m \neq 0$, or $char(K) = 0$ ($1$ has either finite additive order, or infinite additive order).
Clearly, in a finite field, we cannot have characteristic $0$, since we only have finitely many elements to obtain by considering $1,1+1,1+1+1$, etc.
Suppose $char(K) = m$. if m is composite, then $m = st$ for some $1 < s,t < m$.
But a field is an integral domain, so $m1 = 0 \implies (st)1 = (s1)(t1) = 0$, so either $s1 = 0$ or $t1 = 0$, contradicting that $char(K) = m$.
The only two options left then, is that either $m$ is prime, or $m = 1$. but the latter option means $1 = 0$, which is not allowed in a field. So we are forced to conclude that $m$ is prime.
However, Lagrange's theorem applied to the additive group of $K$, tells us the order of $1$ in this group is a divisor of $|K|$, that is: $char(K)|p^n$, hence $char(K) = p$ (this being the only prime that divides $p^n$).