Solve $i^{i^{i^\ldots}}$

Question

How to find $$i^{i^{i^\ldots}} \quad :\quad i=\sqrt{-1}$$ I'm able to find the solution for the finite powers using $$i=e^{i(2k\pi+\frac{\pi}{2})}\quad:\quad k\in\mathbb{Z}$$ $$i^{i}=e^{-(2k\pi+\frac{\pi}{2})}$$ $$i^{i^{i}}=e^{-i(2\pi k+\frac{\pi}{2})}=-i$$ $$i^{i^{i^i}}=e^{(2\pi k+\frac{\pi}{2}) }$$ $$\text{and so on}$$ but what should be the approach to solve for infitie powers$\space$?

Answer 1

$X = i^{i^{i^{\cdot}}}$

So $X = i^X$.

So $\log (X) = X \log (i)$.

So ${\log (X) \over X} = \log (i)$.

So $X = i W(-i)$, where $W$ is the Lambert $W$ or PolyLog function.

Mathematica evaluates this to $X = 0.44 + 0.36 i$.

Here's a graph on the imaginary plane of 200 successive exponentiations with the solution as a red dot: