We can simplify the expression $\sqrt{63}$ by breaking it down into prime factors. We have $63 = 3\cdot 3 \cdot 7$, so $\sqrt{63} = \sqrt{3\cdot 3\cdot 7}= \sqrt{3^2 \cdot 7} = \sqrt{3^2}\cdot \sqrt{7} = 3\sqrt{7}$. Therefore, $\sqrt{63}$ rewritten as the product of an integer and another irrational square root, in simplest form, is $\boxed{3\sqrt{7}}$.
