I was fooling around while waiting for a page to load and came across the following "contradiction".
Let $x=(-1)^{i}$. Then $x^{i}=(-1)^{i\cdot i}=(-1)^{-1}=-1$. Thus, $x=\left(x^{i}\right)^{-i}=(-1)^{-i}=\left((-1)^{i}\right)^{-1}=x^{-1}$. Therefore, $x^2=1$, and so $(-1)^{i}=\pm 1$.
However, we can find a value of $(-1)^i$ that contradicts this. We have $-1=e^{i\pi}$ so that $(-1)^i=e^{-\pi}$, which is certainly not $\pm 1$.
Where is the mistake? I know the answer is absurd and I'm probably being terribly naive about exponents, but it seems solid.
Note that the rule $(a^z)^w=a^{zw}$ is not generally true for complex numbers. As a simple example, using the principal branch for the complex logarithm we see that
$$(e^{i2\pi})^i=1^i=1\ne e^{-2\pi}=e^{i2\pi i}$$
Note that $(1)$ is false since on the principal branch we have
$$\begin{align} x&=(-1)^i\\\\ &=e^{i\log(-1)}\\\\ &=e^{i(i\pi)}\\\\ &=e^{-\pi } \end{align}$$
and
$$\begin{align} (x^i)^{-i}&=\left(-1\right)^{-i}\\\\ &=e^{-i\log(-1)}\\\\ &=e^{-i(i\pi)}\\\\ &=e^{\pi}\\\\ &=1/x\\\\ &\ne x \end{align}$$