Euler's identity confusion: $e^{\pi} = e^{-\pi}$?

Question

We know from Euler's identity that $$e^{\pi i} = -1.$$ So that means $$(e^{\pi i})^{-i} = e^{\pi} = (-1)^{-i} = ((-1)^{-1})^{i} = (-1)^{i}.$$ But also we have $$ (-1)^{i} =( e^{\pi i})^{i} = e^{-\pi}$$ But how can this be, since $e^{\pi} \neq e^{-\pi}$?

Answer 1

You are right about $(-1)^{-i} = (-1)^i$, but it doesn't mean $e^{\pi} = e^{-\pi}$. Please consider the following case as an analogy: $$ (-2)^2 = 4 = (+2)^2, $$ but it does not mean $-2 = +2$. In essence, your equations state $$ (e^{-\pi})^i = (e^{\pi})^i, $$ but it does not imply $e^{-\pi} = e^{\pi}$.