Just use the fact that if,
0<a<b
0<c<d
Then,
0<ac<bd.
Thus, consider "n" inequalities,
0<a<b
0<a<b
.........
0<a<b
Multiply them out,
0<aa...a<bb...b
Thus,
0<a^n<b^n
Or you can prove it by induction (basically what I did).
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