$i^i \approx 1/5$ or what is it? I watched this video by Matt Parker recently: 
https://www.youtube.com/watch?v=9tlHQOKMHGA
He calculates $i^i$ and his answer is around ~1/5
well more precise $e^\frac{-\pi}{2}$ which uses
$e^{i\theta} = \cos(\theta)+i\sin{\theta}$
where $e^{i\frac{\pi}{2}} = i$
so that $i^i = e^{\ln{i^i}} = e^{i\ln{i}} = e^{ii\frac{\pi}{2}} = e^{-\frac{\pi}{2}}$ which is around 1/5
whatever. If you use 
$e^{i\theta} = \cos(\theta)+i\sin{\theta}$
you can also argue that
$e^{i\frac{5\pi}{2}} = i$
and so
$i^i = e^{-\frac{5\pi}{2}}$ 
which is obviously different to $e^{-\frac{\pi}{2}}$.
Where is the error?
WolframAlpha btw agrees with Matt ;)
http://www.wolframalpha.com/input/?i=i%5Ei
 A: There is no error.
There are actually infinite values for $i^i$ due to the periodicity.
A: $i^i$ is by definition $e^{i \log i}$ where $\log z = \log |z| + i \arg z$ is a branch of log.
If you choose your branch of log to be $-\pi < \arg z < \pi$,
then $i^i = e^{-\frac{\pi}{2}}$.
If you choose your branch of log to be $\pi < \arg z < 3 \pi$, then $i^i = e^{-\left(\frac{\pi}{2} + 2 \pi\right)}$
So different branch of log gives you different value of $i^i$.
The branch of log with $-\pi < \arg z < \pi$ is called the principal branch, which is what people usually assumed if the branch of log is not specified. In this branch, the value of $i^i$ is called the principal value of $i^i$, denoted by $\operatorname{PV} i^i$
If you want to know more this kind of stuff, you can take a course in complex number / complex analysis.
A: The secret is on how you define logarithm for complex numbers, if we have a number in the unit circle $x$, we choose its logarithm by convention to be the $0\leq\theta<2\pi$ for which $e^{2\pi i \theta}=x$; otherwise log(1) could be equal to $2k\pi i$ for any integer $k$.
A: There is no error. If $a,b\in\mathbb C$ and $a\neq0$, then $a^b$ stands for any numbers of the form $\exp\bigl(b\log(a)\bigr)$, where $\log a$ can be any logarihtm of $a$.
