Is this an axiom or does it have to be proved?

Question

a and b are positive integers and x is greater than 1. In Rudin principles of real analysis it is not given as an axiom but proving is seems difficult to me

Answer 1

You can prove this by induction:

The base case is $x^{1\cdot b}=x^b=(x^1)^b$.

Then, if $x^{a\cdot b}=(x^a)^b$, $x^{(a+1)\cdot b}=x^{ab+b}=x^{ab}x^b=(x^a)^bx^b=(x^ax)^b=(x^{a+1})^b$ where we used $x^{a+b}=x^ax^b$ and $x^by^b=(xy)^b$.

This proves the identity for natural numbers. After you define the exponents of the form $\frac{1}{n}$ (this will be done in Theorem 1.21), you can expand this to the rational numbers and real exponents will be defined in the exercises (ex. 6).