TITLE: Numbers that are generic w.r.t. exponentiation
QUESTION [42 upvotes]: This is a follow-up to my old question Number of distinct values taken by $x\hat{\phantom{\hat{}}}x\hat{\phantom{\hat{}}}\dots\hat{\phantom{\hat{}}}x$ with parentheses inserted in all possible ways.
In the following let us assume $n$ to be a positive integer, and all other variables to be positive reals. Let $a\hat{\phantom{\hat{}}}b$ denote exponentiation $a^b$. 
The number of distinct $\mathbb{R}^+\to\mathbb{R}^+$ functions obtained from the expression 
$$\underbrace{x\hat{\phantom{\hat{}}}x\hat{\phantom{\hat{}}}\dots\hat{\phantom{\hat{}}}x}_{n\text{ occurences of }x}\tag1$$ 
by inserting parentheses in all possible ways depends on $n$ and is given by the OEIS sequence A000081. Note that different parenthesizations can result in the same function, e.g.
$$(x\hat{\phantom{\hat{}}}x)\hat{\phantom{\hat{}}}(x\hat{\phantom{\hat{}}}x)=(x\hat{\phantom{\hat{}}}(x\hat{\phantom{\hat{}}}x))\hat{\phantom{\hat{}}}x.\tag2$$
If instead of considering functions, we fix some value of $x$, and ask about the number of distinct numeric outcomes of the expression $(1)$ for all possible parenthesizations, then, depending on the value of $x$ we fixed, the result can be either A000081 (in this case we call the value of $x$ generic), or a slower growing sequence. 
For example, the number $2$ is not generic, because the corresponding sequence is A002845 due to some identities specific to the number $2$, e.g.
$$2\hat{\phantom{\hat{}}}(2\hat{\phantom{\hat{}}}2)=(2\hat{\phantom{\hat{}}}2)\hat{\phantom{\hat{}}}2.\tag3$$
Actually, it is not difficult to see that no positive integer is generic. Likewise, $\sqrt2$ is not generic. Furthermore, it can be proved that no positive algebraic number is generic.
Questions:


Can we prove that $2^{\sqrt2}$ is generic?
Can we find an explicit$^*$ computable generic number?


One might think that a plausible candidate for a generic number could be $\pi$, but, unfortunately, we do not even know yet if $\pi^{\pi^{\pi^\pi}}$ is an integer.

$^*$ By explicit I mean something that can be constructed from algebraic numbers and known $^{**}$ constants, elementary and known special functions, or an isolated root of an equation constructed from those constants and functions.
$^{**}$ known means they appeared in published books or reviewed papers.

References:

R. K. Guy and J. L. Selfridge, The nesting and roosting habits of the laddered parenthesis. Amer. Math. Monthly 80 (1973), 868-876. ᵖᵈᶠ
F. Göbel and R. P. Nederpelt, The number of numerical outcomes of iterated powers, Amer. Math. Monthly, 80 (1971), 1097-1103.  ᵖᵈᶠ

REPLY [13 votes]: Two observations:

Not every transcendental is generic. Indeed the real solution $r$ of the equation $x^x=2$ is not generic because $r$ satisfies 
$$x^{x^{x^x}}=\left( x^x\right)^x,$$
and $r$ is transcendental by the Gelfond-Schneider theorem.
Here is an argument showing that no positive real algebraic is generic (by which I mean that every positive real algebraic satisfies an equation between two exponential towers that have the same number of $x$'s and  that do not define the same functions on $\mathbb{R}^+$.) This is mentioned in the original problem but maybe it would be nice to put up a proof here. It seems quite significant to me that 'generic' is a strengthening of 'transcendental'.

Rather than give a formal proof of the claim, I'll present an example that should make the general argument clear. Let $r$ be a solution of the polynomial equation $x^2-3x-2=0$. To prove that $r$ is not generic, first write the equation as
$$xx=x+x+x+1+1.$$
(In general, construct an equation made from additions and multiplications,  all of whose coefficients are 1.)
Equating $x$ raised to the left-hand-side and $x$ raised to the right-hand-side, we obtain the following non-identity (also satisfied by $r$)
$$(x^x)^x=(x^x)(x^x)(x^x)xx.$$
Doing the same thing with the last equation, we get
$$x^{(x^x)^x}=\left(\left(\left(\left(x^{(x^x)}\right)^{(x^x)}\right)^{(x^x)}\right)^x\right)^x.$$
Now call the left and right hand sides of the last equation $a$ and $b$. Then $r$ satisfies the equation $a^b=b^a$, and both $a^b$ and $b^a$ have the same number of $x$'s. Furthermore $a^b$ and $b^a$ cannot define the same function. This follows from the calculus problem to the effect that if $u$ and $v$ are greater than $e$ (the base of the natural logarithm) and if $u^v=v^u$, then $u=v$.