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?
On
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.
On
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.
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.