TITLE: Convergence properties of $z^{z^{z^{...}}}$ and is it "chaotic"
QUESTION [9 upvotes]: $\DeclareMathOperator{\Arg}{Arg}$
Let $z \in \mathbb{C}.$  Let $b = W(-\ln z)$ where $W$ is the Lambert W Function.  Define the sequence $a_n$ by $a_0 = z$ and $a_{n+1} = {a_0}^{a_n}$ for $n \geq 1$, that is to say $a_n$ is the sequence $z, z^z, z^{z^z} ...$
I am trying to classify the sequence $a_n$ as convergent or divergent (i.e. not convergent) according to $|b|$.  My approach to this involves defining $2$ additional sequences: $b_n = \ln a_n$ and $c_n = b_n - b$.  Then $b_{n+1} =  \ln \left(z^{e^{b_n}}\right) = e^{b_n} \ln z$ and $c_{n+1} = e^{(b + c_n)} \ln z - b= (e^b \ln z) e^{c_n}  - b  = b e^{c_n}  - b = b (e^{c_n}  - 1)$.  Suppose $b_n \to b \implies c_n \to 0 \implies e^{c_n} - 1 \sim c_n \implies c_{n+1} \sim bc_n$.  From here there are $3$ cases to consider:
If $|b| > 1$ then $c_n$ is increasing without bound, so it clearly cannot converge except if $c_0 = 0$ which would imply $c_n = 0$ for every $n$ which is not the case because $b_0 =\ln z$ so $b_1 = e^{\ln z}\ln z \ne b_0$ so $c_1 \ne c_0$, a contradiction $\implies b_n$ is divergent $\implies a_n$ is divergent.
If $|b| < 1$ then $c_n$ is decreasing towards $0 \implies b_n$ is convergent $\implies a_n$ is convergent.
If $|b| = 1$ then $c_n \to c \ne 0 \implies b_n$ is divergent $\implies a_n$ is divergent.
I would like to give credit to user1952009.  The method involving the sequences $b_n$ and $c_n$ is due to them.
Edit 1: After reading Yiannis Galidakis's answer several times, I have decided it is mostly satisfactory.  Having said that, I feel I must clarify my position on chaotic sequences.  When I said the concept of a chaotic sequence was nonsense, that was an exaggeration.  What I meant to say was the definition given in the previous post didn't really make sense to me.  The one given here is detailed and precise enough that I can understand what it actually means.
However, I do not believe the statements "$a_n$ is chaotic" and "$a_n$ does not converge" are incompatible.  Setting aside the question of whether $a_n$ is chaotic or not, I feel I have given a plausibility argument (although not a rigorous proof) that $a_n$ is convergent if and only if $c_n \to 0$ which happens if and only if $|b| < 1$.
On a different note, I have been experimenting numerically with $a_n$ for at least $3$ years.  I believe (but cannot prove) that whenever $z$ is not purely real and $|b| = |W(-\ln z)| > 1$ the sequence $a_n$ 
actually has a $k$-cycle for some $k \in \mathbb{N}$.  It appears that $k = 3$ for all sufficiently large $|z|$ and that $k \to \infty$ as $|b| \to 1$.
Similarly, whenever the sequence "shoots off" too a neighborhood of $\infty$ I have found that $a_n \sim$ {$..., \infty, 0, 1, z, z^z ...$}  I will admit however that I cannot predict when the sequence will "shoot of" to $\infty$.
At this point I have $2$ main questions: is my convergence analysis correct?  Are there any values of $z$, not purely real, such that $a_n$ is unbounded, i.e, the sequence "blows up" to $\infty$ and never "comes back down."
Apologies for a somewhat lengthy post.
Edit 2: In Addendum 2 in Yiannis Galidakis's answer, it is stated that if $c \in \mathbb{C}$ and $|\Re(c)| > \exp(\exp(-1))$ and $\Arg{c}=2\pi\alpha$ with $\alpha\in\mathbb{R}\setminus\mathbb{Q}$ then the iterated exponential will be unbounded.  This is not quite correct.  The part about irrational rotation is fine, but it is not enough to have $|\Re(c)| > \exp(\exp(-1))$.  Consider $c = 2\exp(e\pi i/14) \approx 1.6393177+1.1457037i$.  Clearly $c$ satisfies both conditions. However $|W(-\ln c)| \approx .890512$.  Hence the iterated exponential is convergent, hence it is bounded.  A few other counterexamples are given by $c = k\exp(e\pi i/14)$ for $k \in \mathbb{R}$ and $1.76253 \le k \le 2.34896$

REPLY [5 votes]: This is an extended response to Yiannis Galidakis's question.
I have been experimenting with the value $c$ they gave in the previous post.  It is in case 3b since $|t| = |W(-\ln c)| = 1$ and there is no $n \in \mathbb{N}$ such that $t^n = 1$.  I have evaluated the sequence $a_n$ for $0 \leq n \leq 10^8$.  As in Gottfried Helm's analysis, I have found no evidence that the terms are contracting towards some limit.  In fact, it appears the sequence is frequently "slingshot" to a neighborhood of $\infty$, and that the terms are being "slingshot" progressively farther out as $n \to \infty$.  As an example, the largest such "slingshot" has $|a_n| = 17.1$ for $n \leq 10^6$.  For $n \leq 10^8$  this increases to about $25$.  In all such cases, I have observed that $a_n \sim$ {$...,\infty, 0, 1, z, z^z, ...$}.  For these reasons I believe $a_n$ is divergent (i.e. not convergent) in case 3b.
As to whether $a_n$ is chaotic or not, I am not completely sure.  This may be a bad analogy, but, the way I have come to think of it, when we say a sequence or other mathematical object/phenomenon is chaotic, what we're really saying is the condition number of some function is very high relative to some other function.  By this definition, I believe the sequence $a_n$ can be chaotic in all cases, but is especially chaotic in cases 3a and 3b.  For one thing, an arbitrarily small perturbation in $a_0 = z$ can leave the sequence in the same case or shift it into any of the others.  Also, using the formula $cond(f(x)) =  \frac{xf'(x)}{f(x)}$ we see that $a_n$ is very ill-conditioned when it is "slingshot" near $\infty$.
In spite of these issues, I do not lose much sleep over it.  As far as I know, there are $4$ ways to alleviate the difficulty of an ill-conditioned problem.

Supercomputers
Arbitrary precision calculators
A very sophisticated algorithm
A lot of time and effort

Of course there are some problems where even all $4$ of these are not enough.  As someone once said "Computers and calculators are not replacements for mathematical thought. They are just replacements for some kinds of mathematical labor, either numerical or symbolic. There are, and always will be, mathematical problems that can’t be solved by a calculator or computer, regardless of its size and speed."