Why is $\left(e^{2\pi i}\right)^i \neq e^{-2 \pi}$? Here's my (obviously flawed) proof that $1=e^{-2 \pi}$:
$$
1^i=1\\
e^{2 \pi i} = 1\\
\left(e^{2\pi i}\right)^i = 1^i\\
e^{-2 \pi} = 1
$$
What's the issue? I understand that exponentiation is not injective (and thus $-1 \neq 1$ even though $(-1)^2 = 1^2$), but I don't think that's an issue here: I'm only raising things to the power of $i$, which I don't think is multi-valued.

The law of exponents $x^{ab} = (x^a)^b$ does not hold in general. 
Raising numbers to complex powers is multivalued.
For example, $1^i = e^{\log(1^i)} = e^{i (\ln(1)+2\pi k i)} = e^{i(2\pi k i)} = e^{-2\pi k}$ for any integer $k$. Choosing $k=-1$ matches your left hand side, $k=0$ matches the right.
