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.

If $p\in \mathbb{N}$ induction mathematical.
