What is the solution(s) to $1^i$? $1^x$ is always $1$ with real numbers, but everything gets more complicated with complex numbers. Using Eulers formula, you know that 
$$e^{ix}=\cos(x)+i\sin(x)$$
If you make x=2π into this you'd get
$$e^{i2π}=1$$
and raising each side to the power of i gets
$$\left(e^{i2π}\right)^i=1^i$$
$$e^{-2π}=1^i$$
But surely this isn't right? wolfram alpha says this isn't correct, and if it is, could you put in 4π, 6π etc. to get an infinite number of results. 
 A: Since $e^{ix}$ is a periodic function with period $2\pi$, then we have $$e^{2k\pi i} = 1$$ where $k \in \mathbb{Z}$. This means that $$1= e^{2\pi i} = e^{-2\pi i} = e^{4\pi i} = e^{-4\pi i}, \text{etc...}$$
Then raising $1$ to the power of $i$ yields: $$1^{i} = (e^{2k\pi i})^i = e^{2k\pi i^2} = e^{-2k\pi}$$ where $k$ ranges over the integers. 
A: I agree with the comments and the other answer here. The calculation $1^i = (e^{2k\pi i})^i = e^{2k\pi i^2} = e^{-2k\pi}$ for $k \in \mathbb{Z}$ illustrates well how $1^i$ can be considered to be infinitely-valued.
I just wanted to add color to this reasoning a little bit. To do so, let's appeal to a standard definition of exponentiation. For real numbers $x$ and $a$ (with $x > 0$) we typically define $x^a := e^{a\cdot \text{log}(x)}$. Suppose this could be extended seamlessly for $a \in \mathbb{C}$. Then $1^i = e^{i\cdot \text{log}(1)}$. But $\mbox{log}(1)$ is almost surely $0$, so $1^i = e^{0} = 1$. Why does it make sense to have other values?
The root of the issue is the logarithm. As a function $\mathbb{R}_{>0} \rightarrow \mathbb{R}$ the logarithm arises as the inverse of the exponential function $e^x$. It is well-defined on $\mathbb{R}_{>0}$ because the exponential is bijective as a function $\mathbb{R}\rightarrow \mathbb{R}_{>0}$. However, its complex extension $\mbox{$\textbf{exp}$} : \mathbb{C} \rightarrow \mathbb{C}\setminus\{0\}$ is not bijective. In fact, it is not even finite-to-one: it is periodic, with period $2\pi i$. As a consequence, it does not have an inverse function $\mathbb{C}\setminus \{0\} \rightarrow \mathbb{C}$. However if, for example, we restrict the domain of $\textbf{exp}$ to a ribbon $R_a := \{z \in \mathbb{C}: 2a\pi < \mbox{im(z)} < 2(a+1)\pi\}$ (for any $a \in \mathbb{R}$) then $\textbf{exp} : R_a \rightarrow \mathbb{C}\setminus B_a$ is again a bijection (for $B_a: = \{\lambda e^{2a\pi i} : \lambda \in \mathbb{R}_{\geq 0}\}$, often called a 'branch cut' of $\mathbb{C}$), and we can define the inverse map $\mathbb{C}\setminus B_a \rightarrow R_a$, which is often called 'logarithm'. Here though, the logarithm is only a local inverse to the exponential function and there are many different local inverse functions for each choice of $a\in \mathbb{R}$. The actual value for the logarithm of a given complex number will be piecewise-constant with respect to $a$ and will jump by $2\pi i$ when $a$ changes by a full $2\pi$ because the $2\pi i$-periodicity of $\textbf{exp}$ means that $B_a = B_{a+2k\pi}$ if and only if $k \in \mathbb{Z}$ and so we get many different local inverses (parametrized by $k \in \mathbb{Z}$) $B_a \rightarrow R_{a+2k\pi}$ for each individual branch cut $B_a$. Technically, there is no 'correct' choice of local logarithm - when using one, you should just be clear about your choice.
So for any given branch cut $B_a$ (as in the previous paragraph) not containing $1$, there are many different local logarithms $\mathbb{C}\setminus B_a \rightarrow R_{a+2k\pi}$ (one for each $k \in \mathbb{Z}$) and for each one, a different value of $\text{log}(1)$, which ranges over $0 + 2k\pi i$. So the value of $1^i = e^{i\cdot \text{log}(1)}$ ranges over $e^{i\cdot(2k\pi i)} = e^{-2k\pi}$.
