Root of $x^{x^{x^{x^{x^{.^{.^{.}}}}}}} = a$ Root of $x^{x^{x^{x^{x^{.^{.^{.}}}}}}} = a$ where $a > 0$ can be solved by noticing the equation can be transform to $x^a = a$.  Taking the logarithm of base $a$ on the equation, we have $$a \cdot \log_a{x} = 1$$ or $$\log _a x=\frac{1}{a}.$$  This equation has the following solution: $$x=\sqrt[a]{a}.$$  So far it seems to be fine, but notice that both solutions of $$x^{x^{x^{x^{x^{.^{.^{.}}}}}}} = 2$$ and $$x^{x^{x^{x^{x^{.^{.^{.}}}}}}} = 4$$ are $\sqrt{2}$.  What is going on?
 A: If you let $x_0=x,x_{i+1}=x^{x_i}$ then $$x^{x^{x^{x^{x^{.^{.^{.}}}}}}}=\lim_{n\to\infty}x_n$$
may or may not converge. Only if it converges can you write it as $x^a=a$.
A: Let’s make this rigorous. Consider the map $f: x \mapsto x^x$. Iterating this will get you what you want...hopefully.
Choose a compact interval of the real line. Your HYPOTHESIS will be that the limit of the sequence $f(c), f(f(c)), ...$ , where $c$ is in the interval you chose, is some number that you got by your log trick. (In fact, if at all it is that number, there are often conditions in which it HAS to be that number and nothing else).
But to actually prove that the power tower achieves that value, you need to consider the sequence above (function iterates) and show that it converges. 
You should work out for yourself that this logic breaks down where you pointed out exactly because of this failure of the function iteration sequence to converge.
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I also like using the Banach fixed point theorem for these types of problems. 
The reduction of the Banach fixed point theorem to $\mathbb{R}$ is saying: given a compact interval, and a function that maps the compact interval to itself, such that the function is Lipschitz with some constant ratio $\leq 1$ (this is known as a contraction mapping because it shrinks distances), there is a fixed point on that interval and the sequence $f(c), f(f(c)), f(f(f(c)))...$ converges to it for ANY $c$ in that interval.
This is the most well-known fixed point theorem (and the easiest to prove) that says something about the existence of fixed points, an algorithm for calculating them, and a powerful statement about their attractiveness. This last property is what guarantees a finite, convergent number if you take the infinite composition $f(f(f(f...$ of a single number. (Why do you get a fixed number when you whip out your calculator and keep punching cos() for some initial argument? And admit it, you’ve done this.)
This approach might not always work because it’s sometimes hard to find a region where the function is a self map, but if you can easily do this and prove the map is contracting, you can appeal to this theorem without analyzing the behavior of some possibly contrived sequence.
A: Rather than start with $a$ and ask which $x$ satisfies the power tower and possibly converges to $a$ such that $x^a=a$, consider proceeding in the opposite fashion.
Let $1< x \leq 3^{1/3}$ and define $x_n$ recursively by $x_1=x$ and $x_{n+1}=x^{x_n}$ for all $n$. Note that $(x_n)$ is increasing, bounded above and thus by the MCT, converges to a number $a$ such that $a=x^a$, using the continuity of the the exponential function $x^y$ where $x$ is fixed. To see this, use induction. Note $x_2-x_1=x(x^{x-1}-1)\geq 0$ which shows the base case. If $x_n \leq x_{n+1}$ since $x^y$ is an increasing function for $x>1$ we have that $x^{x_n}\leq x^{x_{n+1}}$, i.e. $x_{n+1}\leq x_{n+2}$ (note all $x_n >1$). Clearly $x_1 \leq 3$ and if $x_n \leq 3$ then $x_{n+1}=x^{x_n} \leq x^3\leq (3^{1/3})^3=3$. Thus $(x_n)$ is bounded by $3$. Last, by MCT it has a limit, call it $a$, and we know $a=\lim x_{n+1}=\lim x^{x_n}=x^{\lim x_n}=x^a$ by continuity of $x^y$. So we've tied up any loose ends.
Since $x=\sqrt{2}$ satisfies the very first inequality, we have that the power tower of $\sqrt{2}$ limits to a value $a$ such that $a=\sqrt{2}^a$. The only possible choice is, $a=2$ since we must have $x_n\leq 3$.
A: Another approach is to graph $a$ vs $x^a$, the dependent variable being a and the independent variable being $x$.
Where $x=\sqrt{2}$, there are 2 intersections: where $y=2, 4$. Now let's try a more algebraic approach.
A more general formula is where $y=n$, $x=\sqrt[n]{n}$.
For instance, at $y=3$, $x=\sqrt[3]{3}$.
And here we realize that the answer to your original question is really simple;
At $y=4$, $x=\sqrt[4]{4}$ and at $y=2$, $x=\sqrt{2}$, but $\sqrt[4]{4}$ equals $\sqrt{2}$, so consequently both are solutions for $y$ such that $x=\sqrt{2}$! (Exclamation point mark being a expression of excitement not the factorial or shifted gamma functions)
I thik I owe it to cite my sources, as I did not originally figure out how to do nth root in $\LaTeX{}$, https://www.mathworks.com/matlabcentral/answers/7570-tex-latex-how-to-define-cubic-root
