Proof of strictly increasing nature of $y(x)=x^{x^{x^{\ldots}}}$ on $[1,e^{\frac{1}{e}})$? The title is fairly self explanatory: I have been trying to rigorously prove that $y(x)=x^{x^{x^{\ldots}}}$ is a strictly increasing function over the interval  $[1,e^{\frac{1}{e}})$ for a while now, primarily by exploring various manipulations using logarithms and polylogarithms but have gotten nowhere. Although it is simple enough to show that $y(\sqrt{2})>y(1)$ and if $y'(x)>0$ for some $x \in [1,e^{\frac{1}{e}})$ then $y'(x)>0$ for all $x \in [1,e^{\frac{1}{e}})$ (since either $y$ must be strictly increasing or strictly decreasing), I am not satisfied by the rigor of this argument, although perhaps this is me being too finicky. This lack of progress has led me to explore the possibility that it is only strictly non-decreasing but this loosening of constraints has not helped at all. When it comes to proving that it is a function I've been at a loss as to where I might even begin. Any and all insights are welcome.
 A: We write $$x^{x^{x^{...}}} = \frac{W(-\ln(x))}{-\ln(x)}, x \neq 1$$
We differentiate that. This becomes $$\frac{\ln(x)W'(-\ln(x))+W(-\ln(x))}{x\ln^2(x)} , x \neq 1$$
Using the quotient rule. (W|A verification)
Since $e^{\frac{1}{e}}>x>1$, the numerator is positive. 
Therefore we want to show that 
$$\ln(x)W'(-\ln(x))+W(-\ln(x))>0$$

Let $-\frac{1}{e}<y<0$.
Since $y$ is negative, thus $W(y)$ is negative, we have 
$$0>W(y)$$
$$1>1+W(y)$$
$W(y)>-1$, thus the RHS is postive. Therefore taking the reciprocal will change sign. 
$$1<\frac{1}{1+W(y)}$$
$$-\frac{1}{1+W(y)}+1<0$$
$y$ is negative thus $W(y)$ is negative. Therefore multiplying by it changes sign. 
$$W(y)\left(-\frac{1}{1+W(y)}+1\right)>0$$
$$-y\frac{W(y)}{y(1+W(y))}+W(y)>0$$
$$-yW'(y)+W(y)>0$$
Substitute $y=-\ln(x)$, $-\frac{1}{e}<y<0$, thus $e^{\frac{1}{e}}>x>1$. This gives $\ln(x)W'(-\ln(x))+W(-\ln(x))>0$.
Therefore the derivative is positive so the function is strictly increasing. 
A: It is easier to prove that the inverse function is strictly increasing. Since the inverse function is just:
$$ g(x) = \left(\frac{1}{x}\right)^{-\frac{1}{x}}$$
with a change of variable everything boils down to proving that $h(x)=x^x$ is increasing over $\left[\frac{1}{e},1\right]$. That is trivial since:
$$ h'(x) = h(x)\cdot\frac{d}{dx}\log h(x) = (1+\log x)\,h(x) \geq 0.$$
