Prove if $\displaystyle 0 < a < b\,$ then $\displaystyle 0 < a^n < b^n\,$ for all $\displaystyle n \in \mathbb{N}\,$
observe that this is true for n=1, and then suppose it
true for some positive integer n=k.
Then we have:
0<a<b, and 0<a^k<b^k.
hence as a>0,
0 < a^{k+1} < a b^k
and as b>a:
0 < a^{k+1} < b^{k+1}
Hence by induction if 0<a<b, 0<a^n<b^n for all positive integers n.
RonL