$R$ a $PID \Rightarrow R[[X]]$ a $UFD$

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I have just shown that an integral domain $R$ is a unique factorization domain iff for every non-zero prime ideal $P$ of $R$, $P$ contains a non-zero prime, principal ideal.

I am then asked to show that if $R$ is a $PID$ then $R[[X]]$ (the ring of formal power series over $R$) is a $UFD$.

I have shown that $R[[X]]$ is an integral domain, so it remains to show that if $P$ is a prime ideal of $R[[X]]$, then $P$ contains a principal prime ideal. However, I am not very familiar with working with the ring of formal power series, so I cannot quite see what to do to get the desired result. I would appreciate any help that may be offered, thank you.