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Piero D'Ancona
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I like this one, invented by T.Clausen in 1827: since $e^{2\pi i n}=1$ for all integers $n$, we have $e^{2\pi i n+1}=e$, which implies $e^{(2\pi i n+1)^2}=(e^{2\pi i n+1})^{2\pi i n+1}=e^{2\pi i n+1}=e$. Now expanding the square at the exponent gives $$e^{1-4\pi^2n^2+4\pi n i}=e$$ and after simplifying $$e^{-4\pi^2n^2}=1$$ for all $n$.