complex numbers - Why this proof $0=1$ is wrong?(breakfast joke)

We have $$e^{2\pi i n}=1$$

So we have $$e^{2\pi in+1}=e$$

which implies $$(e^{2\pi in+1})^{2\pi in+1}=e^{2\pi in+1}=e$$
Thus we have $$e^{-4\pi^{2}n^{2}+4\pi in+1}=e$$

This implies $$e^{-4\pi^{2}n^{2}}=1$$

Taking the limit when $n\rightarrow \infty$ gives $0=1$.

Answer

Your error is (as in most of those fake-proofs) in the step where you use the power law $(a^b)^c=a^{bc}$ without the conditions of that power law being fulfilled.