The part you are missing is that we do know the following facts:
- If $n,m\in\mathbb N$, then $(x^{n})^m=x^{nm}$ (which can be shown by induction, as indicated)
- For $n\in\mathbb N$, then $(x^{1/n})^n=x$ (which is by definition of the $n$'th root).
Now we have
$$((b^m)^{1/n})^{nq}=(((b^m)^{1/n})^n)^q=(b^m)^q=b^{mq}=b^{np}=(b^p)^n=(((b^p)^{1/q})^q)^n=((b^p)^{1/q})^{np},$$
then by the uniqueness of the positive root and the fact that $np=mq$, $(b^m)^{1/n}=(b^p)^{1/q}$.
EDIT: To see the first point, we have to understand the definition of $x^n$ for $n\in\mathbb N$. In fact, notice that Rudin does not even bother defining this (probably because it's more technical than required for the purpose of his book).
Let's first look at some example. Consider the multiplication operation on the natural numbers. How is it actually defined? We define it by induction as follows:
$$n\cdot m=n\cdot (m-1)$$ with $n\cdot 1=n$. Note that this actually gives us an algorithm for multiplying two natural numbers. Now, when we want to define $n^p$, for $p$ a natural number, we must also define this inductively as
$$n^p=n^{p-1}\cdot n.$$
Now, with a bit of work, we can extend this to a definition of multiplication on real numbers (Rudin might actually have done this in the appendix, in which he constructs the real numbers), and in the same way define exponentiation similarly as
$$x^p=x^{p-1}\cdot x,$$ and I'll leave it to you to check that $x^p\cdot x^q=x^{p+q}$ (which you can show by induction on $q$).
Now, coming back to the inductive step we see that $$(x^s)^{k+1}=(x^s)^k\cdot(x^s)=(x^{sk})x^s=x^{sk+s}=x^{s(k+1)},$$ as required.
If $S$ is the square with vertices $(0,0),\, (1,0),\, (0,1)$ and $(1,1)$ then $TS=H$, thus
$$\int_{TS} f(x,y) d(x,y)=\int_S(f\circ T)(u,v)|\det\partial T(u,v)| d(u,v)$$
where $\partial T(u,v)$ is the derivative of $T$ at the point $(u,v)\in S$. So yes, you are right, the stated solution is wrong.
Now using Fubini's theorem you can transform this integral over $S$ to a double integral on $[0,1]$ and $[0,1]$.
P.S.: in this case $T$ is an affine map, so it have the form $T=r+A$ for some $2\times 2$ matrix $A$ and some constant $r\in\Bbb R^2$. Hence $\partial T(u,v)=A$ for any chosen $(u,v)\in \Bbb R^2$.
Best Answer
$A_N$ (not $A_n$) is not the set of numbers satisfying the equation $n+\vert a_0\vert+\vert a_1\vert+\cdots+\vert a_n \vert=N$. It is the set of numbers satisfying a polynomial equation $\sum_{i=1}^n a_ix^i=0$ for some coefficients $a_0,\ldots,a_n$ which satisfy the equation $n+\vert a_0\vert+\vert a_1\vert+\cdots+\vert a_n \vert=N$.
And such a polynomial equation has at most $n$ solutions, so each $A_N$ is finite.