What's wrong with the following equations?
$$1 = 1^{-i} = (e^{2πi})^{-i} = e^{-i2πi} = e^{2π}$$
My guess would be the third equation, but I can't really tell why... in the first equation, we use the fact that $1^z$ is still $1$ for any $Z \in C$; in the second step, we insert Euler's Identity... and in the last equation, we simply use $-i2πi = 2π$. But if I'm right with my suspicion, why is the third equation wrong? Can't we use the power rules for complex numbers?
The problem is that complex powers are multi-valued functions. By definition, $a^b = e^{b \log(a)}$, but there are different branches of $\log(a)$, each of which may give a different value to $a^b$. So it's not true in general that $(a^b)^c = a^{bc}$. What you can say is that $(a^b)^c = \exp(c \log(a^b)) = \exp(c \log(e^{b \log a}))$, and $\log(e^{b \log a}) = b \log a + 2 \pi n i$ for some integer $n$, so $$(a^b)^c = \exp(c(b \log a + 2 \pi n i)) = a^{bc} \exp(2 \pi c n i)$$
The possible logarithms of $1$ are $2 \pi n i$ for integers $n$, so $1^{-i} = \exp(-i \log(1)) = \exp(2 \pi n)$ where $n$ can be any integer.
