Demonstration that 0 = 1 [duplicate]
Solution 1:
The problem is that the power rule
$$ (a^b)^c = a^{bc}$$
only holds when $a$ and $b$ are positive real numbers. In that derivation the crucially wrong step is
$$ (e^{2 \pi i n + 1})^{2 \pi i n + 1} = e^{(2\pi i n + 1)(2 \pi i n+1)}.$$
Solution 2:
In complex numbers, $e^a=e^b$ does not imply that $a=b$. For instance, $e^{2\pi in+1}=e$ does not imply that $2\pi in+1=1$.
For the same reason, $\log e^a$ is not the same as $a$, and $(e^a)^a:=e^{(\log e^a)a}$ is not the same as $e^{a^2}$ (instead it is $e^{(a+2\pi ik)a}$, for some $k$).