Lemma 2. Let $R$ be a PID and $p$ a prime element and $a \neq 0$ any element. Then there is an integer $n$ such that $p^n \ | \ a$ but $p^{n+1}$ doesn't divide $a$.
Proof. If the lemma were false, then for each integer $m \gt 0$ there would be an element $b_m$ such that $a = p^mb_m$. Then $pb_{m+1} = b_{m}$ so that $(b_1) \subset (b_2) \subset \dots$ ...
I need help on the part "Then $pb_{m+1} = b_m$". What do they mean?
If both $a = p^nb_n$ and $a = p^{n+1}b_{n+1}$, then $p^nb_n = p^{n+1}b_{n+1}$. The "D" in "PID" is for 'domain,' meaning there are no zero divisors, and thus cancellation holds. So we get that $b_n = pb_{n+1}$ after cancelling $p^n$.
This leads to an infinite ascending chain $(b_i) \subset (b_{i+1}) \subset \ldots$, which cannot happen since PIDs are Noetherian.