Why is $2\pi i \neq 0?$

You have shown that $e^{2\pi i} = e^0$. This does not imply $2\pi i = 0$, because $e^z$ is not injective. You have to give up your intuition about real functions when you move them to the complex plane, because they change a lot. $e^z$ is actually periodic for complex $z$.

The $\log$ function is multi-valued on $\mathbb{C}^*$ (you can however choose a "branch" of it; see Wikipedia). At any rate, just because $$e^{2\pi i}=e^0$$ does not imply that $2\pi i=0$.

It's like saying that, because $\sin{\pi} = \sin{0}$, that $\pi = 0$. Not all functions have perfect inverses; sin being one of them. In the complex numbers, $e^z$ doesn't either.

You're implicitly going: $e^{2\pi i} = e^0 \implies \ln{e^{2\pi i}} = \ln{e^0} \implies 2 \pi i = 0$. The error here is that $\forall x \in \mathbb{C} \ \ln{e^x} = x$ is not true! $ln$ isn't even a function, just like the naive version of $\arcsin(x)$. You have to make a choice of range, which is usually $\Im(x) \in (-\pi, \pi]$.