Proof of strictly increasing nature of $y(x)=x^{x^{x^{\ldots}}}$ on $[1,e^{\frac{1}{e}})$?

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: $$ h'(x) = h(x)\cdot\frac{d}{dx}\log h(x) = (1+\log x)\,h(x) \geq 0.$$

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.