An integral domain I is a nonzero commutative ring in which the product of any two nonzero elements is nonzero.
We need to show that every nonzero element has a multiplicative inverse.
Let be a ∈ I and a ≠ 0.
Consider the mapping:
$$!F: I \rightarrow I\\ \qquad x\mapsto ax$$
We have: ax = ay ⇔ a(x-y) = 0
Since I has no zero divisors and a ≠ 0, we have: x = y.
Therefore: F is injective and since I is finite it is also surjective an bijective.
There is an inverse mapping F-1.
Define a-1 = F-1(1), so we have the existence of the inverse.