I arrive at the following equation:
$\left(-1\right)^a=\left(-1\right)^{-a}$
Intuitively, this equation could be satisfied if $a$ is either $0$ or $1$, however, if you take the log of both sides you get:
$a\ln{\left(-1\right)}=-a\ln{\left(-1\right)}$ $a\left(i\pi\right)=-a\left(i\pi\right)$ $a=-a$
Which can only be satisfied if $a=0$. Why can't $a=1$ since I know that $-1=\frac{1}{-1}=-1$?