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 $-1=\frac{1}{-1}=-1$?