How to prove that $f(x)=x^{x^x}$ is strictly increasing without calculating the derivative?

Question

How to prove that $f:(0,\infty )\to\mathbb R$ defined by $f(x)=x^{(x^x)}$ is strictly increasing without calculating the derivative?

Answer 1

Let $0<x<1<y<z$. Then $$ x^{x^x} < x^0 = 1 = y^0 < y^{y^y} < z^{y^y} < z^{z^z} $$ so $f(x)$ is monotonically increasing in $[1,\infty)$.

Since $f(x)$ is continuously differentiable in $(0,1)$, either it is monotonic in the interval or there exist $w,x\in (0,1)$ with $w<x$ and $f(w)=f(x)$. Suppose such a choice exists and let $$ a = \frac{\log w}{\log x} > 1, \quad w = x^a \\ w^{w^w} = x^{x^x} \\ ax^{ax^a} \log x = x^x \log x \\ \log a + ax^a\log x = x\log x \\ \log a = x \log x (1-ax^{a-1}) $$ Since $x>0$ and $\log x < 0 < \log a$ we must have $$ ax^{a-1}>1 \\ (1-a) \log x < \log a \\ (1-a) \log x < x(1-ax^{a-1}) \log x \\ a-1 < x(ax^{a-1}-1) $$ then either $ax^{a-1}-1<0$ or $x(ax^{a-1}-1)<ax^{a-1}-1<a-1$, either way leading to a contradiction. Hence there can be no such choice of $w,x$. Thus $f(x)$ must be monotonic in $(0,1)$, and it's simple to check that the direction is increasing in $x$.

Answer 2

You want to prove that $\forall\,x_{1},x_{2}\in\mathbb{R},x_{1}<x_{2}\Rightarrow f(x_{1})<f(x_{2})$. For the sake of simplicity let me investigate only the case $1<x_{1}<x_{2}$. We then have $$x_{1}^{x_{1}}<x_{2}^{x_{1}}<x_{2}^{x_{2}}$$ and finally $$x_{1}^{x_{1}^{x_{1}}}<x_{1}^{x_{2}^{x_{2}}}<x_{2}^{x_{2}^{x_{2}}}$$ You can discuss the other possibilities applying appropriate exponentiation properties.