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...

Use induction on $n$:


*

*$n=1$: ok

*$n \to n+1$:  By induction, $x^n \le x$. Multiply by $x \le 1$ and get $x^{n+1} \le x$. Note that $x\ge0$ is essential here.
