What is an easy way to prove the equality $r > s > 0$ implies $x^r > x^s$? I have been using simple inequalities of fractional powers on a positive interval and keep abusing the inequality for $x>1$.  I was just wondering if there is a nice way to prove the inequality in a couple line:
Let $x \in [1,\infty)$ and $r,s \in \mathbb{R}$

What is an easy way to prove the equality $r > s > 0$ implies $x^r > x^s$?


If you accept that $x^y\gt 1$ for $x\gt 1$ and $y \gt 0$, then $x^r=x^{r-s}x^s \gt x^s$ for $x\gt 1$ and $r \gt s$.
