What is wrong with this fake proof $e^i = 1$?

$$e^{i} = e^{i2\pi/2\pi} = (e^{2\pi i})^{1/(2\pi)} = 1^{1/(2\pi )} = 1$$

Obviously, one of my algebraic manipulations is not valid.

As said in the comments, the expressions $(a^b)^c$ and $a^{bc}$ are in fact multivalued functions; they are not a uniquely determined complex number. A classic example of a multivalued function is the complex logarithm denoted by $\log(z)$, $z\in\mathbb{C}$. The complex logarithm $\log(z)$ is any complex number $w$ satisfying $e^w=z$ (which has several solutions, see e.g. this), and hence $\log(z)$ gives rise to a whole set of complex numbers instead of just a single complex number.

Complex exponentiation such as $z^w$ for $z,w\in\mathbb{C}$ is usually defined as $$ z^w=\exp(w\log(z)), $$ where $\log(z)$ is the complex logarithm, and hence this is also a multivalued function. I hope this sheds some light on the problems with doing manipulations on complex numbers as if they were real numbers. See also this for other examples of identities which fail when using complex numbers as they were real numbers.