<s>[INST] Show $\sqrt[\sqrt{i}]{i}\approx 23$. (What are the other values?) Let $\arg(z)=\theta$ and $|z|=r$. So,
\begin{equation*}
\begin{aligned}
\sqrt[\alpha]{z}
& = r^{1/\alpha}\cdot \left(\sin\left(\dfrac{\theta+2\pi k}{\alpha}\right)+i\cos\left(\dfrac{\theta+2\pi k}{\alpha}\right)\right)
\end{aligned}
\end{equation*}
where $k=0,1,\dots,\alpha-1$. 
Using this formula, we want to calculate $\sqrt[\sqrt{i}]{i}$. Note that we will be using this formula twice. Once to calculate $\alpha=\sqrt{i}$ and twice to calculate $\sqrt[\alpha]{i}$.
For $\sqrt{i}$, $r=|i|=1$ and $\theta=arg(z)=\pi/2$. By the formula, for $k=0,1$
$$
\alpha=\sqrt{i}
= 1^{1/2}\cdot \left(\sin\left(\dfrac{\pi/2+2\pi k}{2}\right)+i\cos\left(\dfrac{\pi/2+2\pi k}{2}\right)\right)
$$
For $k=0$, 
$$
\alpha_0=\sqrt{i}
= \sin\left(\dfrac{\pi}{4}\right)+i\cos\left(\dfrac{\pi}{4}\right) =\dfrac{\sqrt{2}}{2} + i \dfrac{\sqrt{2}}{2}
$$
For $k=1$, 
$$
\alpha_1=\sqrt{i}
= \sin\left(\dfrac{\pi/2+2\pi}{2}\right)+i\cos\left(\dfrac{\pi/2+2\pi}{2}\right)
= \sin\left(\dfrac{3\pi}{4}\right)+i\cos\left(\dfrac{3\pi}{4}\right) = \dfrac{\sqrt{2}}{2} - i \dfrac{\sqrt{2}}{2}
$$
For the time being, we will let $\alpha$ act as either $\alpha_0$ or $\alpha_2$ in our computation of $\sqrt[\alpha]{i}$.
\null
For $\sqrt[\alpha]{i}$, $r=|i|=1$ and $\theta=arg(z)=\pi/2$. By the formula, for $k=?$
$$
\alpha=\sqrt{i}
= 1^{1/2}\cdot \left(\sin\left(\dfrac{\pi/2+2\pi k}{\alpha}\right)+i\cos\left(\dfrac{\pi/2+2\pi k}{\alpha}\right)\right)
$$
My question is how am I suppose to use this formula if my $\alpha$'s are complex numbers?
 [/INST] Using the multivalued complex logarithm we have that
$$\frac1{i^{1/2}}=\frac1{\exp(\frac12\ln i)}=\frac1{\exp(\frac12i(\pi/2+2\pi\Bbb Z))}=\frac1{\exp(i(\pi/4+\pi\Bbb Z))}\\=e^{-i\pi/4}e^{i\pi\Bbb Z}=\pm e^{-i\pi/4}=\pm\frac{1-i}{\sqrt 2}$$
Thus
$$i^{1/i^{1/2}}=\exp\left(\pm\frac{1-i}{\sqrt 2}\ln i\right)=\exp\left(\pm\frac{1-i}{\sqrt 2}i(\pi/2+2\pi\Bbb Z)\right)\\=\exp\left(\pm\frac{1+i}{\sqrt 2}(\pi/2+2\pi\Bbb Z)\right)$$
But Im not sure how to simplify this further.
</s>