Rudin assumes $(x^a)^b=x^{ab}$(for real $a$ and $b$) without proof? I am currently self studying Baby Rudin and I'm having some problems with his proof of Theorem 3.20(a) on page 58. I have read all previous chapters and I can't find any mention of real exponents besides its definition in the exercise section of Chapter 1 (but nothing about order/this identity).
He defines $x^a$ for rational $a$ and $x>1$ then this is used to extend it to all real $a$ by $x^a = \sup \{x^t : t \leq a,~ t \in Q\}$.
Am I missing something or does he expect the reader to fill this huge gap in the proof?

Prefatory note on parts for Theorem 3.20: We shall now compute the limits of some sequences which occur frequently. The proofs will all be based on the following remark: If $0\leq x_n\leq s_n$ for $n\geq N$, where $N$ is some fixed number, and if $s_n\to 0$, then $x_n\to 0$.
Theorem 3.20 (a): If $p>0$, then $\lim_{n\to\infty}\frac{1}{n^p}=0$.
Proof. Take $n>(1/\varepsilon)^{1/p}$. (Note that the archimedean property of the real number system is used here.)
 A: First show that it holds for $a, b \in \mathbb N$.
Then show it holds for $a,b \in \mathbb Q$ (this is easy: since $x^{p/q} = x^p \cdot (x)^{1/q}$, and $p, q \in \mathbb N$)
Finally for reals $a, b$ you have that $x^a = \sup\{x^t, t\le a, t \in \mathbb Q\}$; hence you get 
\begin{eqnarray*}(x^{a})^b &=& \sup\{\sup\{x^t, t\le a, t \in \mathbb Q\}^s, s\le b, s \in \mathbb Q\} \\[0.5em]
&=& \sup\{\sup\{(x^t)^s, t\le a, t \in \mathbb Q\}, s\le b, s \in \mathbb Q\} \\[0.5em]
&=& \sup\{\sup\{x^{ts}, t\le a, t \in \mathbb Q\}, s\le b, s \in \mathbb Q\} \\[0.5em]
&=& \sup\{x^{ts}, t\le a, s\le b, s \in \mathbb Q, t \in \mathbb Q\} \\[0.5em]
&=&\sup\{x^{k}, k\le ab, k \in \mathbb Q\} \\[0.5em]
&=& x^{ab}
\end{eqnarray*}
A: "Am I missing something or he expects the reader to fill this huge gap in the proof?" It's not only a gap in the proof, Rudin hasn't even defined what $1/n^p$ means for real $p>0.$ In fact beyond integer powers, he has only defined $a^p$ for $a\ge 0$ and $p=1/n$ for some $n\in \mathbb {N}.$ Yes, there are the exercises in Chapter 1 concerning rational and real powers, but the whole lot of them make for a huge expenditure of effort at that point. I don't think this is one of Rudin's finest moments.
A: Well, when we consider integers...
Note that
$$
(x^a)^1 = x^a = x^{a \cdot 1}
$$
Use induction:
IF $(x^a)^{n} = x^{an}$, then also $(x^a)^{n+1} = x^{a(n+1)}$ , as
$$
(x^a)^{n + 1} = (x^a)^n x^a = x^{an} x^a = x^{a(n+1)}
$$
It is true for $n=1$, and by induction it is true for any positive integer.
