Prove that if $a > b > 0, p > 0$, then $a^p > b^p$ 
Prove that if $a > b > 0, p > 0$, then $a^p > b^p.$

As I was reading baby Rudin, this fact was a step that Rudin skipped (Theorem 3.20a), but it is not obvious to me how to prove this.
Thanks in advance.
EDIT (relevant definitions and results from exercise 6, chapter 1): 
EDIT 2 - Presume $b > 1$.
Let $r = m/n, n>0$, where $m$ and $n$ are integers. Then $b^r = (b^m)^{1/n}$.
It is proved that $b^{r+s} = b^rb^s$. If $x$ is real and if we let $$B(x) = \{  b^r | r \in \mathbb{Q}, r\leq x \}$$
then $b^r = sup  B(r)$ and we define $b^x = sup B(x) $ for any real $x$.
Also it is proved that $b^{r+s} = b^xb^y$, for real $x$ and $y$.
Hopefully that helps, my bad for not including it the first time around.

You are using Rudin's "Principals of Mathematical Induction" and you are doing Chapter 1, Excercise 6.  Which relies very heavily on the Theorem 1.21 and the proof thereof  that;
For any $b > 1$ and $n \in \mathbb N$ there is a unique positive $c$ so that $c^n =b$.  We call such a $c:= b^{\frac 1n}$.
The proof makes use of the least upper bound property and the archimedian principal and the fact that for all $c < b$ then $c^n < b*c^{n-1} < b^2*c^{n-2} < ....< b^n$.  We then consider $C= \sup \{c|c^n < b\}$ and... the proof writes itself.
But HERE's the thing.  In the process of doing this we have established that for all $c < b$ than so that $d= c^n < b$ that $c < b^{\frac 1n} = \sup \{c|c^n < b\}$.  Thus for $d < b$ we have $c = d^{\frac 1n} < b^{\frac 1n}$.
And if that WASN't immediately clear, it'd have to be by contradiction:
If $d^{\frac 1n} \ge b^{\frac 1n}$ then $d=(d^{\frac 1n})^n \ge (b^{\frac 1n})^n = b$ which is a contradiction.
So if we show that it is consistent to define for $p =\frac mn$ that $b^p = (b^{\frac 1n})^m$ we would have $0 < a < b  \iff 0 < a^{\frac 1n} < b^{\frac 1n} \iff a^{\frac mn} < b^{\frac mn}$.
And it'd only take a line to extend that result to $a^x = \sup \{a^q|q< x; q\in \mathbb Q\}< \sup\{a^q|q< x; q\in \mathbb Q\} = b^x$.
Which is why Rudin "took it for granted".
====in recap  ==
For natural numbers it's clear by induction.
If $a^n > b^n > 0$ then $a^{n+1} =a^n*a > a^n*b > b^n*b = b^{n+1}$.
For $p = \frac 1n; n\in \mathbb N$ it's clear by contradiction.
If $a^{\frac 1n}\le b^{\frac 1n}$ we'd have $a = (a^{\frac 1n})^n \le (b^{\frac 1n})^n = b$.  
So for rational $p = \frac nm; n,m\in \mathbb n$ then $a^p = (a^n)^{\frac 1m} > (b^n)^{\frac 1m} = b^p$.
ANd for any real $x>0$ we have $a^x = \sup\{a^q|q < x\} > \sup \{b^q|q < x\}$ [admittedly that step would need a sentence or two but it'd be straight forward] $= b^x$. 
