Complex Exponential False "Proof" That All Integers Are $0$

The following false "proof" is attributed to Thomas Clausen in 1827, and was stated on page 79 of Nahin's An Imaginary Tale.

$e^{i2\pi n}=1$ for all integers $n$. So

\begin{align*} ee^{i2\pi n}=e&=e^{1+i2\pi n}\\ &=\left(e^{1+i2\pi n}\right)^{1+i2\pi n}\\ &=e^{(1+i2\pi n)^2}=e^{1+i4\pi n-4\pi^2n^2}\\ &=e^{1+i4\pi n}e^{-4\pi^2n^2} \end{align*} But $e^{1+i4\pi n}=e$, therefore $e^{-4\pi^2n^2}=1$. But that last equation is only true for $n=0$. We started with a statement true for all integers $n$, and through a series of (apparently) valid steps ended with a statement true only for $0$. Therefore all integers are $0$. Where is the mistake?

Solution 1:

See Failure of power and logarithm identities and Complex exponents with positive real bases 2 which explains this exact problem. Also consider the related problem of $e^{-4\pi^2n^2}=\left(e^{i2\pi n}\right)^{i2\pi n}=1^{i2\pi n}=1$, which forgets that, in this scheme, the power of 1 is involves a multivalued log of 1 in an exponent.