indeed. I hope you understood the steps leading upto $\displaystyle p|b^2$.ok..i understand where you're coming from now. so since p / b^2 it also divides b..

The definition of prime-numbers say: if $\displaystyle p|mn \Rightarrow p|m$ or $\displaystyle p|n$

Apply that idea to $\displaystyle p|b\cdot b$ then follows that $\displaystyle p|b$.