What is the maximum convergent $x$ in the power tower $x^{x^{x^{x\cdots}}}$?

Fix an $x > 0$. Since the map $a \mapsto x^a$ is continuous, we know that if the infinite tower $x^{x^{x^\cdots}}$ converges to some limit $y$ then $x^y = y$, which implies $x = y^{1/y}$. Elementary calculus shows that the maximum possible value of $y^{1/y}$ occurs at $y = e$, so it is impossible for the infinite tower to converge unless $x \le e^{1/e}$.

It remains to show that the sequence actually converges for $x = e^{1/e}$. To prove this we need only establish the inequality $1 < y \le x^y \le e$ for any $y$ in the range $1 < y \le e$. Then a simple induction will show that the sequence $x, x^x, x^{x^x}, \ldots$ is increasing and bounded, hence convergent. Finally, the desired inequality is easy to prove using the aforementioned calculation that shows $e^{1/e}$ is the unique maximum of the function $y^{1/y}$.

Example 4 of the Wikipedia page on the Lambert W function tells how to solve $x^y=y=x^{x^{x^{x\cdots}}}$ for $y$ using the function: $$x=y^{\frac1y}$$ $$\frac1x=y^{-\frac1y}=\left(\frac1y\right)^{\frac1y}$$ $$-\ln x=\frac1y\cdot\ln\frac1y=\ln\frac1y\cdot e^{\ln\frac1y}$$ $$W(-\ln x)=\ln\frac1y$$ $$\frac1y=e^{W(-\ln x)}=\frac{-\ln x}{W(-\ln x)}$$ $$y=\frac{W(-\ln x)}{-\ln x}$$ The largest possible value of $x$ that will make this expression for $y$ defined over the reals satisfies $-\ln x=-\frac1e$, where the RHS is the lower limit of the domain of the real-valued Lambert W. Therefore the maximum convergent value of $x$ is $$e^{\frac1e}=1.444667861\dots$$