Why is $\underbrace {i^{i^{i^{.^{.^{.^{i}}}}}}}_{n \ times}$ converging? Introduction:
Recently I found out that $i^i \approx 0.20788$ has no imaginary part. I got interested and then wanted to know whether there are other $n$ for which $\underbrace {i^{i^{i^{.^{.^{.^{i}}}}}}}_{n \ times}$ has no inaginary part.
So I wrote this python script (I'm a beginner at python, could be very bad code :) which plots what I call the $i-Tower\ up\ to\ n = 100$. It looks like this:

Question:
Let's use this convention: ${}^ni = \underbrace{i^{i^{i^{.^{.^{.^{i}}}}}}}_{n \ times}$.


*

*Why is the sequence ${}^ni\ |\ n \in \mathbb{N}$ converging?
${}^{20}i \approx 0.48770 + 0.41217i$ 
${}^{60}i \approx 0.437584 + 0.360535i$ 
${}^{100}i \approx 0.43829+ 0.36059i$
What I already noticed is that the angle between the lines you can draw from ${}^ni$ over ${}^{n+1}i$ to ${}^{n+2}i$ is $<90°$.


*

*Is that angle equal for all $n$?

*Is there a way to give the exact value of $x = {}^ni \ $with$\ {n \rightarrow \infty}$?

*What is the connection between $i$ and $x$. Is $x$ a special complex number that is maybe already known or comes up in other places?
 A: *

*The infinite power tower converges to the value:


$$i^{i^{i^{\cdot^{\cdot^{\cdot^{\cdot}}}}}}=-\frac{\text{W}(-\ln(i))}{\ln(i)}=\frac{2\text{W}\left(-\frac{\pi i}{2}\right)i}{\pi}\approx0.438286+0.3605924i\tag1$$
See: MathWorld


*

*Solving this equation:


$$a^b=b\Longleftrightarrow b=-\frac{\text{W}(-\ln(a))}{\ln(a)}\tag2$$
With $\text{W}(z)$ is the product log fuction


Notice, for the 'Product Log Function' (or the Lambert $\text{W}$-function) is defined as follows:
$$f(z)=ze^z\to z=f^{-1}(ze^z)=\text{W}(ze^z)\tag3$$

Now, we know that for $k\in\mathbb{R}^+$ (real, and bigger than zero) $\ln(k)$ is well defined.
Now, your question is about:
$$y(k)=k^{k^{k^{k^{\dots}}}}=-\frac{\text{W}\left(-\ln(k)\right)}{\ln(k)}\tag4$$
Eisenstein's (1844) considered this series of the infinite power tower. $y(k)$ converges iff $e^{-e}\le k\le e^{\frac{1}{e}}$; OEIS A073230 and A073229), as shown by Euler (1783) and Eisenstein (1844) (Le Lionnais 1983; Wells 1986, p. 35).
So, the domain of $y(k)$:
$$\left[0,e^{\frac{1}{e}}\right]\tag5$$
