This is from Rudin Exercise 1.6(a).
It asks to prove that if $m, p \in \mathbb{Z}$ and $n,q \in \mathbb{N}$ satisfy $\frac{m}{n} = \frac{p}{q}$ then for all 0 < b $\in \mathbb{R}$, prove that $(b^m)^\frac{1}{n} = (b^p)^\frac{1}{q}$
I know we can show that any $x^{\frac{1}{n}}$ exists and is unique from a previous proof by Rudin, but I'm completely stuck. Any form of help would be much appreciated!
Suppose \begin{align} \frac{m}{n} = \frac{p}{q} \ \ \Rightarrow \ \ mq = np \end{align} then it follows \begin{align} b^{mq}= b^{np} \end{align} because $mq$ and $np$ are integers. Next, by existence/uniqueness of radicals, we have \begin{align} (b^m)^q =b^{mq} = b^{np} \ \ \Rightarrow \ \ b^m = (b^{np})^{1/q} = (\underbrace{b^p \cdots b^p}_{n-\text{times}})^{1/q} = \underbrace{(b^p)^{1/q}\dots (b^p)^{1/q}}_{n-\text{times}} = [(b^p)^{1/q}]^n. \end{align} Thus, again by the existence/uniqueness of radicals, we have that \begin{align} (b^{m})^{1/n} = (b^p)^{1/q}. \end{align}