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...
 A: For $0 < x \le 1$,
$$ \dfrac{d}{dn} x^n = \dfrac{d}{dn} \exp(n \log(x)) = \log(x) \exp(n \log(x)) \le 0$$
because $\log(x) \le 0$ while $\exp(n \log(x)) > 0$.
Do the case $x=0$ separately.
A: Let
$g(n) = x^n
$.
$g(n)
= e^{n \ln(x)}
$,
so
$g'(n)
=\ln(x) e^{n \ln(x)}
=\ln(x) x^n
$.
Since,
if $0 < x < 1$,
$\ln(x) < 0$
and
$x ^ n > 0$,
$g'(n)
< 0 $
so $g(n)$
is decreasing for
$0 < x < 1$.
Therefore,
if $n > 1$,
$g(n) < g(1)$
or
$x^n < x$.
A: Note that $a \leq b$ and $c \geq 0$ implies $ac \leq bc$. One way to see this: we have $(b-a)c \geq 0$, as a product of two non-negative numbers, hence $bc \geq ac$.
Now, assuming $0 \leq x \leq 1$ we may use the above rule and find $x^2 \leq x$. Repeat as long as necessary: $x^3 \leq x^2$, then $x^4 \leq x^3$, and so on. Thus, for every integer $n \geq 1$ we have
$$ x^n \leq x^{n-1} \leq x^{n-2} \leq \cdots \leq x^2 \leq x \leq 1.\tag*{(1)} $$

Edit: I failed to read the comments; you were asking about real values of $n$, not just integers. The other answers show how to apply techniques from calculus to prove the general result. So I'm going to be stubborn and attempt to salvage my algebraic solution. This is by no means the easiest solution.
The next step is to prove the result for rational values of $n$, using the result for integer values of $n$. Write $n = \frac{a}{b}$ with $a,b\in\mathbb{Z}$ and $a \geq b \geq 1$. Note that $x^{\frac{1}{b}} \leq 1$ must hold, for otherwise we would have
$$ x \: = \: \underbrace{x^{\frac{1}{b}}\cdots x^{\frac{1}{b}}}_{b\ \text{times}} \: > \: 1, $$
as a product of $b$ real numbers each strictly larger than $1$. This is a contradiction, for we assumed $x \leq 1$, so it follows that $x^{\frac{1}{b}} \leq 1$ holds, as promised. Of course we also have $x^{\frac{1}{b}} \geq 0$, no matter which construction is used to define $x^{\frac{1}{b}}$. Hence by (1) we have
$$ x^n = x^{\frac{a}{b}} \leq x^{\frac{a-1}{b}} \leq \cdots \leq x^{\frac{b+1}{b}} \leq x \leq x^{\frac{b-1}{b}} \leq \cdots \leq x^{\frac{2}{b}} \leq x^{\frac{1}{b}} \leq 1.\tag*{(2)} $$
This proves that the inequality $x^n \leq x$ holds for all rational $n \geq 1$.

Finally, to extend the result to all real $n > 1$, we have to use some continuity. Specifically, we use that the map $f_x : \mathbb{R}_{>1} \to \mathbb{R}$ given by $f_x(n) = x^n$ is continuous. Now let $n\in\mathbb{R}_{>1}$ be given. Choose some sequence $\{m_k\}_{k=1}^\infty$ of rational numbers in $\mathbb{R}_{>1}$ that converges to $n$. By continuity we have
$$ x^n = \lim_{k\to\infty} x^{m_k}. $$
Each term of this sequence lies in the closed interval $[0,x]$, so the limit must also lie in that interval. In particular, we have $x^n \leq x$.
A: If $p > 0$ then the function $f(t) = t^p$ is increasing on $[0,\infty)$. You can prove this using the first derivative.  Thus if $0 \le x \le 1$ then $0 \le x^p \le 1$.
If $n > 1$ then $p - 1 > 0$ so that
$$ 0 \le x \le 1 \implies 0 \le x^{n-1} \le 1 \implies 0 \le x^n \le x.$$ 
A: 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.
A: 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 $x^n\le x$.
