Artinian implies IZ

Verbal statement
Any Artinian ring is IZ: every element is either invertible, or a zero divisor.

Proof
Given: An Artinian ring $$A$$, and an element $$x \in A$$

To prove: $$x$$ is invertible or a zero divisor

Proof: Consider the descending chain of ideals:

$$A \supset (x) \supset (x^2) \supset \ldots$$

By the Artinianness, this chain stabilizes at some point, so we have:

$$x^n = ax^{n+1}$$

for some $$a \in A$$. Rewriting, we see that:

$$x^n( 1 - ax) = 0$$

If $$1 - ax = 0$$, then $$x$$ is invertible. Otherwise, $$x^n$$ is a zero divisor, and hence $$x$$ is a zero divisor.