Taking the imaginary to the imaginary

By Euler's formula , $e^{\theta i} = \cos \theta + i \sin \theta$ , we have $\large \blue i^i = \blue e^{\blue{\left(2n\pi + \frac \pi 2\right)i}\times i} = \boxed{e^{2n\pi -\frac \pi 2}}$ , where $n$ is an integer.

First of all, note that $i^i = e^{i\ln(i)}$

$\ln(i)$ is tricky, but by using Euler's formula $\left(e^{i\theta}=\cos\theta + i\sin\theta\right)$ we can determine that $e^{i\frac{\pi}{2}}=\cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right)=0+i(1)=i$

Knowing this, we see that $\ln(i)=\ln\left(e^{i\frac{\pi}{2}}\right)=i\frac{\pi}{2}$ and we can conclude that $i^i=e^{i\left(i\frac{\pi}{2}\right)}=e^\frac{-\pi}{2}$

However, this is not the only answer because Euler's formula works $\frac{\pi}{2}$ plus all positive and negative multiples of $2\pi$ (think about rotating all the way around a circle to end back at the start position, in this case $\frac{\pi}{2}$ )

Therefore we can finally conclude that, having $n$ be a constant that exists in the set of all positive and negative integers:

$i^i=e^{i\left(i\frac{\pi}{2}+2i\pi n\right)}=\boxed{e^{2\pi n - \frac{\pi}{2}}, n\in\Z}$

Edit: Didn't include the $2\pi n$ inside $\theta$