Every irreducible is product of irreducibles

Question

I'm studying some commutative algebra and learning some of the bases. While taking a look at the proof of Theorem 1, proposition 2, i've stumbled upon the following: "Suppose $d$ is not a product of irreducible elements. Hence, $d$ isn't irreducible". I'm trying to understand this statement, and it seems pretty trivial, but my head is just not working anymore. If anyone could help, I'd appreciate.

Theorem 1: given $D$ an integral domain.

1. If $D$ is and UFD, and $\pi \in D$ is an element such that $\pi \neq 0$ and $\pi \not\in D^{\times}$, then $\pi$ is prime if, and only if, $\pi$ is irreducible;

2. Suppose that

   (i) Every ideal of $D$ is finitely generated (i.e. $D$ is Noetherian);

   (ii) Every irreducible element is a prime element.

   Then, $D$ is an UFD.

Answer 1

Important: In this contexts, products of irreducibles can consist of one factor only!

Thus if $a$ is irreducible, then $a$ is an irreducible product.

Do you see why the statement is trivial then?

Compare with the integers: Every non-unit integer is the product of primes. Here every prime $p$ is a trivial product. The statement would no longer be true if we did not allow such products (with one factor) and that's definitely not something we want.