prove or disprove that $x^{x^x}$ is one to one on the interval $(0,\infty)$

First, $\log\bigl(x^{x^x}\bigr)=x^x\log x$. We want to show this is increasing for $x>0$.

Differentiate it to get $x^x(1/x+\log x+(\log x)^2)$. Now $(\log x)^2\geq 0$, and $x^x> 0$ for $x>0$.

Also, $e^y>y$ for any $y$, and consequently $e^{1/x}>1/x$ for $x>0$, or equivalently $1/x>\log 1/x=-\log x$.

Putting all this together, $\frac{d}{dx}\log\bigl(x^{x^x}\bigr)>0$.

We also need $\lim_{x\to 0^+}x^{x^x}=0$; this is true because $\lim_{x\to 0^+}x^x=1$.