Infinite powering by $i$ [duplicate]

Find the value of:

$i^{i^{i^{i^{i^{i^{....\infty}}}}}}$

Simply infinite powering by i's and the limiting value.

Thank you for the help.

Let us denote $x=i^{i^{i^{i^\cdots}}}$. Then we have $$i^x=x.$$ It looks like the solution is $x= \frac{2i}{\pi} W(-i\pi/2)$ with $W$ Lambert's $W$ function. Now, $W$ is multivalued. You have to figure out which of the different branches $x$ converges to (and if it converges at all). Numerically, you find (using the principal branch of the logarithm to define the exponentiation) that $x= 0.438283 + 0.360592 i$ which corresponds to the principal branch.

Knowing that you should be able to prove the result by some kind of fixed point theorem.

Here is a numerical result supporting Fabian's argument.

Here, the complex logarithm

$$ z^{w} := \exp (w \operatorname{Log} z) $$

is defined via the principal value $\mathrm{Log}$ of the logarithm, defined on $\Bbb{C} \setminus (-\infty, 0]$.

Another way is to take natural logarithms:

$$i^{i^{i^{i^{i^{i^{\dots \infty}}}}}}=y$$

$$\ln y= \ln (i)^y$$

$$y\ln i=\ln y$$

$$\ln i=\dfrac{i \pi}{2}$$

$$\dfrac{y.i\pi}{2}= \ln y$$

$$e^{\frac{iy\pi}{2}}=y$$

This is rather another comment than an answer but contains a picture, so...
If we display the 3-step-like trajectory separated into 3 single trajectories, we get an improvement of imagination of the convergence. See this one

Similar improvements can be made with other bases. The idea is, to use this for convergence-acceleration procedures like Euler-sums and similar.

[Update]: Also the process of convergence can be improved over the need to iterate 100 times and more. Just use the Newton-iteration. Here is a code-snippet in Pari/GP:

f(x)  = exp( L *x)    \\ implements x->  b^x where L is the log of te base b
fd(x) = L * exp(L*x)  \\ implements the derivative of f(x)

L = log(I)
x0=0.5+0.5*I       \\ Initialize
[x0=x0 - (f(x0)-x0)/(fd(x0)-1)  , exp(L*x0)-x0]  \\ repeat this, say, 7 times

Result:

x0=0.5+0.5*I    \\ initialize
 %214 = 0.500000000000 + 0.500000000000*I

[x0=x0 - (f(x0)-x0)/(fd(x0)-1)  ,  exp( L*x0)-x0]   \\ repeat this say 7 times
 %215 = [0.429683379978 + 0.358463904092*I, 0.0149144114062 - 0.00263680525658*I]
 %216 = [0.438282449555 + 0.360624709917*I, -0.0000214307236671 - 0.0000508331490807*I]
 %217 = [0.438282936547 + 0.360592471486*I, 0.000000000547853619231 + 0.000000000479209718138*I]
 %218 = [0.438282936727 + 0.360592471871*I, 1.24483565546 E-19 - 2.36342583549 E-20*I]
 %219 = [0.438282936727 + 0.360592471871*I, -1.59860647096 E-39 - 3.49116795082 E-39*I]
 %220 = [0.438282936727 + 0.360592471871*I, 2.79037134755 E-78 + 2.15595352591 E-78*I]
 %221 = [0.438282936727 + 0.360592471871*I, 2.83277459577 E-156 - 9.05172112238 E-157*I]
 %222 = [0.438282936727 + 0.360592471871*I, 5.10320381 E-203 - 2.551601908 E-203*I]
  \\ convergence sufficient, 200 dec digits