Every finite integral domain is a field

Let $a \in D$ be any nonzero element, and consider the left-multiplication map $L_a : D \to D$ defined by $L_a(x) = ax$.

We first show $L_a$ is injective. Suppose $L_a(x) = L_a(y)$, so $ax = ay$, i.e. $a(x - y) = 0$. Since $D$ is an integral domain and $a \neq 0$, there are no zero divisors, which forces $x - y = 0$ and hence $x = y$.

Because $D$ is a finite set, any injective map $D \to D$ is automatically surjective (by the pigeonhole principle: $|L_a(D)| = |D|$). In particular, the multiplicative identity $1 \in D$ must lie in the image of $L_a$, so there exists $b \in D$ with $L_a(b) = ab = 1$. Thus $b = a^{-1}$, and since $a \neq 0$ was arbitrary, $D$ is a field.