$b^{r+s}=b^r b^s$

Question

Rudin, Principles of Mathematical Analysis, 3e, page 22, problem 6b.

Fix $b>1$. Prove that $b^{r+s}=b^r b^s$, where $r$ and $s$ are rational numbers.

Below is the solution to the above problem taken from this link. Unfortunately, I don't understand how it is justified to write that $b^{mq+np}=b^{mq}b^{np}$. I'd appreciate a clarification.

Answer 1

It is justified because $mq, np$ are both integers. We assume that we know $b^{r+s}=b^rb^s$ already, for $r,s$ integers.

Answer 2

For integers, $b^n$ is defined as $b\cdot b\cdots b$, where there are $n$ $b$s. Using that multiplication is associative, one can prove the additive property of exponents, which allows you to conclude what you wish to.