i raised to itself Last night I learned the amazing fact that $i^i=e^{-\pi/2}$
So I started computing other powers and would like to get confirmation that:
$i^{i^i}=e^{-i\pi/2}$
$i^{i^{i^{i}}}=e^{\pi/2}$
$i^{i^{i^{i^{i}}}}=e^{i\pi/2}$
Then the pattern continues
(I tried an online calculator to confirm ${i^i}^{i^i}$ but got an imaginary part, and a real part that didn't match)
Now I'm reminded of those infinite power towers you see on Twitter and wondering what would happen if you took $i^{i^{i^{i^{i^{.{^{.^{.}}}}}}}}$ 
I tried writing $z = i^{i^{i^{i^{i^{.{^{.^{.}}}}}}}}$ then rewriting this as $z=i^z$ thus $z^{1/z}=i$ but not sure where to go from here, and I'm not even sure if what I'm doing makes sense. Any thoughts?
 A: Actually there are infinit solutions:
$$
\begin{align*}
z &= \mathrm{i}^{\mathrm{i}}\\
z &= (\mathrm{i})^{\mathrm{i}}\\
z &= (\mathrm{e}^{\ln(\mathrm{i})})^{\mathrm{i}}\\
z &= (\mathrm{e}^{(\frac{\pi}{2} + 2 \cdot k_{k \in \mathbb{Z} \cdot \pi}) \cdot \mathrm{i}})^{\mathrm{i}}\\
z &= \mathrm{e}^{(\frac{\pi}{2} + 2 \cdot k_{k \in \mathbb{Z} \cdot \pi}) \cdot \mathrm{i} \cdot \mathrm{i}}\\
z &= \mathrm{e}^{(\frac{\pi}{2} + 2 \cdot k_{k \in \mathbb{Z} \cdot \pi}) \cdot \mathrm{i}^{2}}\\
z &= \mathrm{e}^{(\frac{\pi}{2} + 2 \cdot k_{k \in \mathbb{Z} \cdot \pi}) \cdot (-1)}\\
z &= \mathrm{e}^{-(\frac{\pi}{2} + 2 \cdot k_{k \in \mathbb{Z} \cdot \pi})}\\
z &= \mathrm{e}^{-\frac{\pi}{2} - 2 \cdot k_{k \in \mathbb{Z} \cdot \pi}}\\
\\
\text{only if } k = 0 \text{:}\\
z &= \mathrm{e}^{-\frac{\pi}{2} - 2 \cdot k_{k \in \mathbb{Z} \cdot \pi}}\\
z &= \mathrm{e}^{-\frac{\pi}{2} - 2 \cdot 0 \cdot \pi}\\
z &= \mathrm{e}^{-\frac{\pi}{2}}\\
\end{align*}
$$
It makes a diffrens when you calculate
$$  i^{(i^{i})} \text{ or } (i^{i})^{i} $$
'cause
$$
\begin{align*}
\mathrm{i}^{(\mathrm{i}^{\mathrm{i}})} &\text{ or } (\mathrm{i}^{\mathrm{i}})^{\mathrm{i}}\\
\mathrm{i}^{\mathrm{e}^{-\frac{\pi}{2} - 2 \cdot k_{k \in \mathbb{Z} \cdot \pi}}} &\text{ or } \mathrm{i}^{\mathrm{i} \cdot \mathrm{i}}\\
\mathrm{i}^{\mathrm{e}^{-\frac{\pi}{2} - 2 \cdot k_{k \in \mathbb{Z} \cdot \pi}}} &\text{ or } \mathrm{i}^{\mathrm{i} \cdot \mathrm{i}}\\
\mathrm{i}^{\mathrm{e}^{-\frac{\pi}{2} - 2 \cdot k_{k \in \mathbb{Z} \cdot \pi}}} &\text{ or } -\mathrm{i}\\
\end{align*}
$$
$$ \begin{align*}
z &= \mathrm{i}^{\mathrm{i}^{\mathrm{i}^{.{^{.^{.}}}}}}\\
z &= \mathrm{i} \uparrow\uparrow \infty\\
z &= \mathrm{i}^{z} \quad\mid\quad (\text{ })^{\frac{1}{z}}\\
z^{\frac{1}{z}} &= \mathrm{i} \quad\mid\quad (\text{ })^{-1}\\
(z^{\frac{1}{z}})^{-1} &= \mathrm{i}^{-1}\\
z^{-1 \cdot \frac{1}{z}} &= \frac{1}{\mathrm{i}}\\
(z^{-1})^{\frac{1}{z}} &= \frac{\mathrm{i}}{\mathrm{i}^{2}}\\
(\frac{1}{z})^{\frac{1}{z}} &= \frac{\mathrm{i}}{-1}\\
(\frac{1}{z})^{\frac{1}{z}} &= -\mathrm{i}\\
&\Rightarrow z = \frac{\operatorname{W}(\ln(\frac{1}{i}))}{\ln(\frac{1}{\mathrm{i}})}\\
z &= \frac{\operatorname{W}(\ln(\frac{1}{\mathrm{i}}))}{\ln(\frac{1}{\mathrm{i}})}\\
z &= \frac{\operatorname{W}(\ln((-\frac{\pi}{2} + 2 \cdot k_{k \in \mathbb{Z} \cdot \pi}) \cdot \mathrm{i})))}{\ln((-\frac{\pi}{2} + 2 \cdot k_{k \in \mathbb{Z}} \cdot \pi) \cdot \mathrm{i})} \quad\mid\quad \text{let us use } k = 0\\
z &= \frac{\operatorname{W}(\ln((-\frac{\pi}{2}) \cdot \mathrm{i})))}{\ln((-\frac{\pi}{2}) \cdot \mathrm{i})} \quad\mid\quad \text{let us use the main branch of the lambert W-Function}\\
z &= 0.438282936727032111626975163551264824267897351646394603609221240... +
0.360592471871385485952940526906000653826577030786027004741451298... \cdot \mathrm{i}
\end{align*}
$$
PS
A nice calculator for stuff like this is Wolfram|Alpha...
A: Complex exponents aren't well-defined. Some exponents with a non-real base and a non-real exponent yield infinitely many different values. In fact, over the complex numbers, even $1^\pi$ can take on infinitely many different values depending on how many whole turns $1$ is considered to be.
