prove or disprove that $x^{x^x}$ is one to one on the interval $(0,\infty)$ I have looked at the graph of $x^{x^x}$ and it appears to be one to one on the interval $(0, \infty)$. I think it should be one to one. Can somebody please prove this or disprove this?
 A: 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 $\lim_{x\to 0^+}x^x=1$.
A: The function $f$ is smooth, maps $(1,\infty)$ into itself, and $(0,1)$ into itself, maps $1$ to itself. 
Besides, $f \rightarrow 0$ at $0$ and $f \rightarrow \infty$ at $\infty$, thus $f$ is surjective. 
If $x >1$, $\ln(\ln{f(x)})=\ln(x^x\ln{x})=x\ln{x}+\ln(\ln{x})$ thus $f$ is increasing on $(1,\infty)$. 
If $0 < x <1$, $R(x)=\ln(-\ln{f(x)})=\ln(\ln{x})+x\ln{x}$. 
The derivative of $RHS$ is $\frac{1}{x\ln{x}}+\ln{x}+1$. 
Now, if $u > 0$, $\frac{e^u}{u}+u \geq 2e^{u/2} > 1$ by AM-GM. Therefore $R’(e^{-u}) < 0$, thus $R’ <0$ on $(0,1)$ and $f$ increases on $(0,1)$. 
Thus $f$ is one-to-one.
A: Let $f:(0,\infty)\to\mathbb{R}$ be defined by
$$f(x)=x^{x^{\Large{x}}}$$
To show that $f$ is one-to-one, it will suffice to show that $f$ is strictly increasing.

Computing $f'$, we get
$$f'(x)=f(x)x^{x-1}\bigl(x\ln(x)^2+x\ln(x)+1\bigr)$$
hence to show $f$ is strictly increasing, it suffices to show $f'(x) > 0\;$for all $x > 0$.

Since


*

*$f(x) > 0$, for all $x > 0$.$\\[4pt]$

*$x^{x-1} > 0$, for all $x > 0$.


it remains to show 
$$x\ln(x)^2+x\ln(x)+1 > 0$$ 
for all $x > 0$.

Consider $3$ cases . . .

Case $(1)$:$\;x\in [1,\infty)$.
\begin{align*}
\text{Then}\;\;&x\in [1,\infty)\\[4pt]
\implies\;&\ln(x)\ge 0\\[4pt]
\implies\;&x\bigl(\ln(x)^2+ln(x)\bigr)\ge 0\\[4pt]
\implies\;&x\ln(x)^2+x\ln(x)+1\ge 1\\[4pt]
\end{align*}
so case $(1)$ is done.

Case $(2)$:$\;x\in \bigl(0,{\large{\frac{1}{e}}}\bigr]$.
\begin{align*}
\text{Then}\;\;&x\in \bigl(0,{\small{\frac{1}{e}}}\bigr]\\[4pt]
\implies\;&\ln(x)\le -1\\[4pt]
\implies\;&\ln(x)^2\ge |\ln(x)|\\[4pt]
\implies\;&\ln(x)^2+\ln(x)\ge 0\\[4pt]
\implies\;&x\bigl(\ln(x)^2+\ln(x)\bigr)\ge 0\\[4pt]
\implies\;&x\ln(x)^2+x\ln(x)+1\ge 1\\[4pt]
\end{align*}
so case $(2)$ is done.

Case $(3)$:$\;x\in \bigl({\large{\frac{1}{e}}},1\bigr)$.

Noting that the quadratic function 
$$t\mapsto t^2+t$$
has minimum value $-\frac{1}{4}$, it follows that
$$\ln(x)^2+\ln(x) \ge -{\small{\frac{1}{4}}}$$
for all $x > 0$, hence
\begin{align*}
&x\in \bigl({\small{\frac{1}{e}}},1\bigr)\\[4pt]
\implies\;&x\bigl(\ln(x)^2+\ln(x)\bigr)\ge -{\small{\frac{1}{4}}}\\[4pt]
\implies\;&x\ln(x)^2+x\ln(x)+1\ge {\small{\frac{3}{4}}}\\[4pt]
\end{align*}
so case $(3)$ is done.

This completes the proof.
