Where is wrong on my $i^i=1$ proof? I wanted to calculate $i^i$ ($i$ is an imaginary unit), and I calculated in this way:
$i^i = (i^4)^{i/4} = 1^{i/4} = 1$ (because $1^x = 1$ for any number $x$)
So, I thought that $i^i$ 's one solution is $1$.
Though, in Wolfram Alpha, it says the solution was $e^{2n\pi - \pi/2}$. It doesn't include $1$. What's wrong with my proof?
 A: There are some errors. In first place in general $a\neq (a^n)^{1/n}$, otherwise we will had that $-1=((-1)^2)^{1/2}=1$.
By the other hand the equation $a=x^n$ have $n$ distinct solutions for $x$ (some times one or two of them are real). Then, in general, we define $x^z:=e^{z\ln x}$, where $\ln$ is understood as the principal value of the complex logarithm.
A: One of the central questions of complex variable is what is the logarithm?  You skirt this blithely.  Pay attention! Open the book by Ahlfors on Complex Variable.
A: For $z,w\in\mathbb C$, we define exponentiation by $z^w\equiv \exp(w\ln z)$, where here $\ln$ is the complex logarithm. Now it might seem like this doesn't rule out your solution. After all, the exponentiation rules for logarithms says that $n\ln z = \ln(z^n)$. But the complex logarithm is a multi-valued function, and thus any expression involving it is only well-defined once we choose a branch. If we stick to the principal branch (i.e., the one where $|\mathrm{Im}(\ln z)|\le \pi$), then we have to modify the rule to
$$
n\ln z = \ln z^n \;\;\;\mathrm{if} \;|\mathrm{Arg}(z)|\le \pi/n
$$
as otherwise the left hand side would be on a different branch of the complex logarithm.
This then resolves your problem. $|\mathrm{Arg}(i)| = \pi/2 > \pi/4$, so you can't apply the rule.
