Is $(-1)^x$ multivalued?

Question

$(-1)^\pi=(e^{i\pi})^\pi=e^{i\pi^2}=\cos(\pi^2)+i\sin(\pi^2)$. Wolfram Alpha lists this as the only answer. However it started with $e^{i\pi}=1$, although $e^{i\pi(2n+1)}=-1$ also for any integer n. By substituting this new expression for $-1$ and doing the same thing, you get more than one value, like $\cos(3\pi^2)+i\sin(3\pi^2)$. Is Wolfram or my reasoning wrong?

Answer 1

$(-1)^x=e^{(1+2n)\pi ix}$ which has infinite possibilities for all irrational $x$.