TITLE: Rudin assumes $(x^a)^b=x^{ab}$(for real $a$ and $b$) without proof?
QUESTION [12 upvotes]: 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.)

REPLY [5 votes]: "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.