Euler's Formula ($e^{ix}=\cos(x)+i\sin(x)$) is pretty useful here. Since $\cos((2n+1)\pi)=-1$ and $\sin((2n+1)\pi)=0$ for integer $n$, $$\cos((2n+1)\pi)+i\sin((2n+1)\pi)=e^{i(2n+1)\pi}=-1$$
$$ (-1)^z = (e^{i(2n+1)\pi})^z = \cos((2n+1)\pi z)+i\sin((2n+1)\pi z) \tag{$n\in\mathbb{Z}$} $$ (This equation can also be viewed as a Generalization of De Moivre's formula)
Since $n$ goes over all integers, $(-1)^z$ has _infinite_ values if the exponent $z$ is irrational. If $z$ is rational, its values repeat after a certain value of $n$.
Take $(-1)^\pi$ as an example: $$ \begin{align} (-1)^\pi &= (e^{i(2n+1)\pi})^\pi\\\ &=e^{i(2n+1)\pi^2} \\\ &=\cos((2n+1)\pi^2)+i\sin((2n+1)\pi^2) \tag{$n\in\mathbb{Z}$} \end{align} $$