What is a clever way to show that if $0 \leq x \leq 1$ then $x^n \leq x$ for every $n > 1$ ? ($n \in \mathbb{R}$) What is a clever way to show that if $0 \leq x \leq 1$ then $x^n \leq x$ for every $n > 1$ ? ($n \in \mathbb{R}$)
I tried to do it with derivatives but I didn't manage to show why this is true...

If $x=0$, $x^n=x$. If $0<x\le1$, $\log x \le0$, so for $n>1$, $n\log x\le1\log x$, and because $e^x$ is an increasing function, $e^{n\log x}\le e^{1\log x}$. But $e^{n\log x}=x^n$ and $e^{1\log x}=x$, so $e^{n\log x}\le e^{1\log x}$ (just shown to be true) is just another way to write $x^n\le x$.
