Finite integral domain is a field

Multiplication by a nonzero element $a$ of an integral domain is injective, by multiplicative cancellation. When the domain is finite, this self-map is surjective, so some element $b$ satisfies $ab=1$. Commutativity gives $ba=1$, so every nonzero element has a multiplicative inverse and the domain is a field.

Even when the domain convention does not initially assume an identity element, finiteness produces one. Surjectivity supplies $e$ with $ae=a$; then $a(ex-x)=0$ forces $ex=x$ for every $x$. Commutativity makes $e$ a two-sided identity, after which the inverse argument applies.