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 $0=1$.