If $a>0$, $b>0$ and $n\in \mathbb{N}$, show that $aIf $a>0$, $b>0$ and $n\in \mathbb{N}$, show that $a<b$ if and only if $a^n<b^n$.  Hint: Use mathematical induction.
Having trouble with the proof that if $a<b$ then $a^n<b^n$.
So far I have;
Assume $a<b$
then $a^k<b^k$ for $k=1$
Assume $\exists m \in \mathbb{n}$ such that $a^m<b^m$
Then let $k=m+1$ so $a^{m+1}<b^{m+1}$
Then $a*a^m<b*b^m$
I'm not positive I can make the second assumption and if I can I don't know how to prove the last statement which would allow the extension into all natural numbers.

It's clear for $k=1$. Suppose it is true for $k=n$ then
$a^{n+1}=a\cdot a^{n}<a\cdot b^{n}<b\cdot b^{n}=b^{n+1}$.
Note that we used that $a>0$ and $b>0$ to preserve direction of inequality. That is we first used the induction hypothesis $a^{n}<b^{n}$ and then since $a>0$ we have:
$a^{n+1}=a\cdot a^{n}<a\cdot b^{n}$.
Then using that $a<b$ and that $b^{n}>0$ since $b>0$ we get:
$a\cdot b^{b}<b\cdot b^{n}=b^{n+1}$.
