$\ e^{i 2\pi} = 1,$  $\ e^{0} = 1$  $\Rightarrow $  $\ e^{i 2\pi} = e^{0}$  $\Rightarrow $  $\ {i 2\pi} = 0 $  $\Rightarrow $  $\ i = 0 $
We know that $e^{\pi i} = -1$ because of de Moivre's formula. ($e^{\pi i} = \cos \pi + i\sin \pi = -1).$  Suppose we square both sides and get $e^{2\pi i} = 1$(which you also get from de Moivre's formula), then shouldn't $2\pi i=0$? What am I missing here?
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