Prove by induction that, for all natural numbers, if 0 < x < y, then x^n < y^n.
I write out the most imporant step in this solution but you need to complete all the steps of the proof thyself. If $\displaystyle 0<x<y$ and $\displaystyle x^n < y^n$ then it means $\displaystyle x\cdot x^n < x\cdot y^n$. Therefore, $\displaystyle x^{n+1} < x\cdot y^n$ however, $\displaystyle x\cdot y^n < y\cdot y^n = y^{n+1}$. Putting this together we find that $\displaystyle x^{n+1} < y^{n+1}$.