If $a>0$, $b>0$ and $n\in \mathbb{N}$, show that $a

If $a>0$, $b>0$ and $n\in \mathbb{N}$, show that $a

If $a>0$, $b>0$ 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.