How do you solve $(i^{i})^{i}$? I am trying to solve this problem, and show that it does not necessarily equal $i^{-1}$. My method so far is
$$(i)^{i} \\
=e^{(i)\ln{(i)} }
i = \cos(\theta)+i\sin(\theta)
\Rightarrow \theta = \frac{\pi}{2}\pm2n\pi, n \in \mathbb{Z} \\
\Rightarrow i = e^{i\frac{\pi}{2}\pm2in\pi} \\
\Rightarrow \ln{(i)} = i\frac{\pi}{2}\pm2in\pi \\
\Rightarrow e^{(i)\ln{(i)} } = e^{(i)(i\frac{\pi}{2}\pm2in\pi)}\\
=e^{(-\frac{\pi}{2}\pm2n\pi)}$$
However when I raise this to the power of $i$ once more, the answer ends up becoming $i^{-1}$ once more. What am I missing here? If someone can please show me where I'm going wrong that would be fantastic!
 A: This question is a duplicate of a previous one, however the answers does not really explain in detail what is going on with respect to OPs calculations so I'll add a CW here to try address this.
First let's define properly what we mean by $(i^i)^i$. The standard way to define this is by taking $z^a \equiv e^{a\text{Log}_n(z)}$ where $\text{Log}_n(z)$ is the $n$'th branch of the logarithm defined by
$$\text{Log}_n(z) = \log|z| + i\arg(z) + 2\pi i n\tag{1}$$
We then have
$$(i^i)^i \equiv e^{i\,\text{Log}_n(e^{i\,\text{Log}_n(i)})}$$
Now $i\,\text{Log}_n(z) = -\frac{\pi}{2} - 2\pi n$ is real however the logarithm $\text{Log}_n$ of the real number $e^{i\,\text{Log}_n(z)}$ does not have to be. This is where the mistake is in your calculation. Using $(1)$ above we see that
$$\text{Log}_n(e^{i\,\text{Log}_n(i)}) = -\frac{\pi}{2} - 2\pi n + 2\pi in$$
so
$$(i^i)^i \equiv e^{-\frac{\pi i}{2} - 2\pi i n - 2\pi n} =  \frac{e^{-2\pi n}}{i}$$
Only for the principal branch $n=0$ is this equal to $\frac{1}{i}$.
