Baby Rudin exercise 1.6: Is this the proof Rudin expects? 
$\bf Exercise\, 1.6$
Fix $b>1.$
Prove that if $m,n,p,q$ are integers, $n>0,q>0$ and $r=m/n=p/q$, then
$$
(b^m)^{1/n}=(b^p)^{1/q}.
$$

I'm not really sure what I can assume and what  I can't assume, I think that all I need is $(x^y)^z=x^{yz}$ for integers $y,z$, but I'm not sure how to prove this (I don't even know what the expected definition of exponentiation is!).
Attempt
$$
\begin{align}
\left((b^m)^{1/n}\right)^n&=b^m\quad \text{By Theorem 1.21 (I think).}\\
\left((b^m)^{1/n}\right)^{nq}&=b^{mq}\quad \text{Here I use $(x^y)^z=x^{yz}$.}\\
\left((b^m)^{1/n}\right)^{nq}&=b^{np}\quad \text{As $mq=np$.}\\
\left((b^m)^{1/n}\right)^{qn}&=b^{pn}\quad 
\end{align}
$$
Then, taking $n$-th and then $q$-th roots, we get our desired result (I think this is possible, again, by theorem $1.21$, but I'm not sure).
Could someone check my proof and tell me which facts about exponentiation we are allowed to assume and use for these kind of proofs?
 A: In general, Rudin seems to assume we know about integers but not reals.  Because of this, it looks like 
(a) the definition of $b^i$, with $i$ a positive integer, is inferred from 1.13 ($i$ copies of $b$ multiplied together, just like for the case of $b$ an integer), 
(b) $(b^i)^j=b^{ij}$ can be assumed known for positive integers $i, j$ ($j$ copies of $b^i$ multiplied together are the same as $ij$ copies of $b$ multiplied together, say "by associativity"), 
(c) similar things can be done for $i$ and $j$ negative, with $b^i$ being $-i$ copies of $1/b$ multiplied together,
(d) from the Corollary to Theorem 1.21 we can also infer $(b^i)^{1/n}=(b^{1/n})^i$ for positive integers $i$ by induction on $i$:  if we assume $(b^i)^{1/n}=(b^{1/n})^i$ already shown for some $i$, let $a=b^i$ in the Corollary; we then get $(b^{i+1})^{1/n}=(b^ib)^{1/n}=(b^i)^{1/n}b^{1/n} =(b^{1/n})^i b^{1/n} = (b^{1/n})^{i+1}$.  From here we can also get the result for $i$ negative.
Assume wlog that $p$ and $q$ have no common divisor.  From our background knowledge regarding integers, we know that $m=jp$, $n=jq$ for some integer $j$.  
Our goal is $(b^m)^{1/n}=(b^p)^{1/q}$, that is $(b^{jp})^{1/jq}=(b^p)^{1/q}$. Because of the uniqueness of $n$th roots in Theorem 1.21, we just need to show that $((b^{jp})^{1/jq})^q=b^p$.  Based on (d) above, $$((b^{jp})^{1/jq})^q=(b^{jpq})^{1/jq}=((b^p)^{jq})^{1/jq}=b^p$$ as desired.
It's so much fun to prove stuff with one hand tied behind your back!
A: I'll assume that at this point you have seen the definition for $x^n$ and $x^{1/n}$ where $n\in\mathbb{N}$. The point of the exercise is to show that we can define $x^r$ for rational $r$ in such a way that doesn't depend on any particular representation of $r$.
Let $z=(b^m)^{1/n}$. It's enough to show that $z^q=b^p$. By definition, we have
$$
z^n=b^m\implies z^{nq}=b^{mq}=b^{pn}\implies z^q=b^p
$$
because both equal to the unique $n$-th root of the common value of $z^{nq}=b^{pn}$. Here, we use Theorem 1.21.
The above is very similar to your argument. The difference is in the observation that $z^n=b^m$ doesn't need Theorem 1.21.
A: Rudin defined $b^n; n \in \mathbb Z$ as notation to mean $b^n = b\cdot b ..b$.  Simple grouping allows you to assume $b^nb^m = b \cdot b...b \cdot b\cdot b....b = b^{n+m}$ and $(b^n)^m = b^{nm}$.
By theorem 1.21 you know that for $b^m$ there exists a unique $d := (b^m)^{1/n}$ such that $d^n = b^m$.
The excercise is to show if $m/n = p/q$ then $(b^p)^{1/q} = (b^m)^{1/n}$.
Your proof is mostly good.
$((b^p)^{1/q})^{mq} = ([(b^p)^{1/q})^q]^m = ([b^p]^m) = b^{pm}$  So $(b^p)^{1/q} = (b^{pm})^{1/mq}$ which is a uniquely defined number by th 1.21
$((b^m)^{1/n})^{np} = ([(b^m)^{1/n})^n]^p = ([b^m]^p) = b^{pm}$  So $(b^m)^{1/n} = (b^{pm})^{1/np}$ which is a uniquely defined number by th 1.21
But $1/np = 1/mq$ so $(b^{pm})^{1/np} =(b^{pm})^{1/mq}$ are both the same unique number.
So $(b^p)^{1/q} = (b^{pm})^{1/mq} = (b^{pm})^{1/np}= (b^m)^{1/n}$ are all different ways of expressing the same unique number.
Thus defining $b^r$ as $(b^m)^{1/n}$ when $r = m/n$ is consistant, not ambiguous, and always existent.  Thus it is "well-defined".
