Prove that $(b^m)^{\frac{1}{n}} = (b^p)^{\frac{1}{q}}$ The exercise is:

Fix $b>1$. If $m, n, p, q$ are integers, $n>0$, $q>0$ and $r=m/n=p/q$, prove that:
$$(b^m)^{\frac{1}{n}} = (b^p)^{\frac{1}{q}}$$
Hence it makes sense to define $b^r = (b^m)^{\frac{1}{n}}$

So far, this is what I got:
Since $n,q>0$, $m$ and $p$ must both be positive or both be negative, or else the equality $m/n=p/q$ won't hold. Now, $m/n$ is not necessarily an integer, but $mq$ surely is. The same holds for $np$. It is trivial to see that $(b^m)^{\frac{1}{n}} = (b^p)^{\frac{1}{q}} \iff b^{mq} = b^{np}$.
Lemma: For every real $x>0$ and every integer $w>0$, there is one and only one positive real $y$ such that $y^w = x$.
Proof: See Rudin's Principles of Mathematical Analysis, page 10.
Hence, $mq=np$. So, indeed $(b^m)^{\frac{1}{n}} = (b^p)^{\frac{1}{q}}$.
Is this proof satisfactory? Am I missing something?
 A: I think you have the right idea.  One minor comment is that when you write “Hence, $mq=np$”, you imply that the statement $mq=np$ follows from some immediately previous statement, such as the the lemma.  But it doesn't; it follows from the hypothesis that $m/n = p/q$.  
But more importantly (IMO), this line

It is trivial to see that $(b^m)^{\frac{1}{n}} = (b^p)^{\frac{1}{q}} \iff b^{mq} = b^{np}$

needs more explanation.  When students write that something is “trivial”, I usually take it as a bluff.  To me the point of the exercise is to use the lemma you cited to prove that $y^n = b^m \implies y^q = b^p$.  Your highlighted statement follows.

Addendum: Based on the comments below, I think you're still missing the point of the exercise and how the lemma is to be used.  You're right that the lemma guarantees $(b^m)^{1/n} = (b^p)^{1/q}$, but if I understand the exercise correctly, you are to provide the details.
The lemma tells us that every positive number has a $w$-th root, and how to recognize the $w$-th root of a number from its properties.  In other words, $w$-th roots exist and are unique.
What is $(b^m)^{1/n}$? It's the unique positive real number $y$ such that $y^n = b^m$.  Likewise, $(b^p)^{1/q}$ is the unique positive real number $z$ such that $z^q = b^p$.  To prove that $(b^m)^{1/n} = (b^p)^{1/q}$, it is sufficient to show that $y^q = b^p$ Why? And how?
A: "" It is trivial to see that $(b^m)^{\frac{1}{n}} = (b^p)^{\frac{1}{q}} \iff b^{mq} = b^{np}$""
Whether that is or is not trivial, that is the one thing you are asked to show and you must show it.  You did not.
You "lemma" is Rudin's Th. 1.21 and you need to use it twice.
Here's a hint:  Use the theorem to prove that $(b^m)^{\frac 1n}$ and $(b^p)^{\frac 1q}$ exist in the first place. Let $(b^m)^{\frac 1n} = d$ and $(b^p)^{\frac 1q} = c$.  Find an integer $k$ so that $d^k = c^k> 1$.  Use the theorem to prove $d = k$.
