Nested powers of $\sqrt 2$ has a solution different from its limit. What does this mean? The infinitely nested power expression below has a limit of $2$:
$$x=\sqrt2^{\sqrt2^{\sqrt2^{...}}}$$
In finding this limit, we may use:
$$x=(\sqrt2)^x$$
But this expression has two solutions, $2$ and $4$.
We know that $2$ is the right answer by evaluating some finite truncations, but this $4$ is bothering me. What does $4$ mean in this expression? Is it significant in some way?
 A: The problem here is that:
$$x = (\sqrt{2})^x$$
is a necessary condition but not sufficient, that is, if $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\cdots}}}$ equals $x$ then it must hold that $x = (\sqrt{2})^x$, but you don't know if $x$ can equal  $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\cdots}}}$ in the first place.
One interesting property is that
$$x^{x^{x^{x^{\cdots}}}}$$ for positive x converges only if $x\in [e^{-e}, e^{1/e}]$ and converges to  a value $y\in[e^{-1}, e]$.
So if you are working with another expression like that and you get more than one solution, keep in mind that if it does not belong to that interval then it surely is not a solution of the expression.
https://www.maa.org/programs/maa-awards/writing-awards/exponentials-reiterated-0
Here is the proof of the property aforementioned and also, a more detailed analysis of the whole problem with nested power expressions.
A: If $\sqrt{2}^{\vdots}$ is well-defined, it must be as the limit of a sequence of the form $a_{n+1}=\sqrt{2}^{a_n}$. The limit, if it exists, depends on $a_1$, but the most natural choice of $a_1$ is $\sqrt{2}$; what else do we think is "at the top of the tower", which doesn't exist? 
But we can prove by induction that if $a_1\in[0,\,2]$, then, for all $n$, $a_n\le 2$, so $x:=\lim\limits_{n\to\infty}a_n$ is $x=2$, whereas $a_1=4$ implies $x=4$.
A: It is a common misconception that an expression like 
$$\sqrt2^{\sqrt2^{\sqrt2^{...}}}$$
denotes a well-defined number.
For more rigor, let us use
$$x=\lim_{n\to\infty}a_n,\text{ where }x_{n+1}=\sqrt2^{x_n}.$$
Now if $x_0=2$, $x_n=2\ \forall n$ follows. Similarly, for $x_0=4$, $x_n=4$ follows. For other initial values, the sequence may converge to $2$, but may also diverge.
So the statement "has a limit of $2$" is dubious.
