Euler showed that:
$$ e^{i\pi} + 1 = 0 $$
So I thought:
$$ b \in \mathbb R,i = \sqrt{-1} \\ n^{bi} = e^{bi\ln n}\\ b = 2k\pi\\ e^{bi\ln n} = e^{2k\pi i\ln n} = (e^{i\pi})^{2k\ln n} = ((-1)^2)^{k\ln n} = 1 $$ I'm not sure that all operations are valid.