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.

Here's another way round,
Since $b^n - a^n = (b-a)[a^{n - 1} + a^{n - 2}b + ... + b^{n - 1}]$ $\cdots(1)$
Given $(1)$ is $> 0$ also for $a>0$ and $b>0$
$[a^{n - 1} + a^{n - 2}b + ... + b^{n - 1}]>0$
hence $b>a$.
