Finite Integral Domain is Galois Field/Proof 2

Theorem
A finite integral domain is a field.

Proof
Let $R$ be a finite integral domain with unity $1$ and zero $0$.

Let $R^*$ denote $R \setminus \left\{{0}\right\}$, the set $R$ without the zero.

As $R$ is finite, we may enumerate the elements of $R$ as:
 * $x_0 = 0, x_1 = 1, x_2, x_3, \ldots, x_n$

Let $x_k \in R$ such that $x_k \ne 0$.

Consider the elements:
 * $x_k x_1, x_k x_2, \ldots, x_k x_n$ which are certainly members of $R$ by closure.

We have:
 * $x_k x_p = x_k x_q \implies x_p = x_q$ by the cancellation property for integral domains.

So the list of products consists of $n$ distinct elements and is therefore equal to the whole $R^*$.

Since $1 \in R^*$, we have that $x_k x_{k^*} = 1$ for some $k^*$.

Therefore, there exists an inverse for every non-zero element, so $R$ is a field.