$R$ UFD and integral domain $\Rightarrow$ every $a\neq 0 \in R\setminus R^\times$ is a product of prime elements

Question

Let $a$ be as in title. Then by definition of UFD we have unique factorisation $$a=\epsilon p_1\dots p_n \quad : \epsilon\in R^\times \text{ and } p_i \in R \text{ irreducible}.$$ It is known that in an UFD irreducible elements are prime. To finish the proof we have to argue $\epsilon = 1$, but why is that?