Element of Principal Ideal Domain is Finite Product of Irreducible Elements

Theorem
Let $R$ be a principal ideal domain.

Let $p \in R$ such that $p \ne 0$ and $p$ is not a unit.

Then there exist irreducible elements $p_1, \ldots, p_n$ such that $p = p_1 \cdots p_n$.

Proof
If $p$ is irreducible, it is proven.

Suppose $p$ is not irreducible, then $p = r_1 r_2$ where neither $r_1$ nor $r_2$ are units.

If $r_1$ and $r_2$ are irreducible, then it is proven.

If $r_1$ is not irreducible then $r_1 = r_{11} r_{12}$ where neither $r_{11}$ nor $r_{12}$ are units.

If we continue, we get a proper inclusion of ideals $\left({r}\right) \subset \left({r_1}\right) \subset \left({r_{11}}\right) \subset \cdots \subset R$.

If this process finishes in a finite number of steps, the proof is complete.

With a view to obtaining a contradiction, suppose that the process does not complete in a finite number of steps.

Thus the chain $\left({r}\right) \subset \left({r_1}\right) \subset \left({r_{11}}\right) \subset \cdots \subset R$ is infinite.

From Principal Ideal Domain fulfills Ascending Chain Condition, this cannot happen.

Thus, the process ends in a finite number of steps.