In a field $F$, the non-zero elements form a group with respect to multiplication. Thus if a multiplicative inverse exist , it must be unique.
It is easy to show that any integral domain must have characteristic either $0$ or a prime $p$.
i.e. $(1+1+...+1)=0\implies$ if $char F= mn\,,\,m,n<char F$ then $(m\cdot 1)(n\cdot 1)=0$ so either $m\cdot 1=0$ or $n\cdot 1=0$ . That is $char F=m$ or $n$. So $char F$ is not composite. i.e. it is a prime.
Now for any finite field $F$ of characteristic $p$, you have an surjective homomorphism from $f:\Bbb{Z}\to F$ given by $f(n)=n\cdot 1$.
Then the kernel is $(p\Bbb{Z})$ and hence any finite field contains $\frac{\Bbb{Z}}{(p\Bbb{Z})}=\mathbb{Z}_{p}$.
Then $F$ forms a vector space over $ \mathbb{Z}_{p}$ and it must be finite dimensional as $F$ is finite.
So say it has dimension $n$ with basis $\{a_{1},a_{2},...,a_{n}\}$
Then $$F=\{\sum_{i=1}^{n}c_{i}a_{i}:c_{i}\in \mathbb{Z}_{p}\}$$
So there are $p$ choices for each $c_{i}$ and hence $|F|=p^{n}$.
So any finite field is of cardinality $p^{n}$ for some prime $p$ and positive integer $n$.