False proof that $ρe^{iθ} = ρ$ and so complex numbers do not exist?

Question

My professor showed the following false proof, which showed that complex numbers do not exist. We were told to find the point where an incorrect step was taken, but I could not find it. Here is the proof: (Complex numbers are of the form $\rho e^{i\theta}$, so the proof begins there) $$\large\rho e^{i\theta} = \rho e^{\frac{2 \pi i \theta}{2\pi}} = \rho (e^{2\pi i})^{\frac{\theta}{2\pi}} = \rho (1)^{\frac{\theta}{2\pi}} = \rho$$ $$Note: e^{i\pi} = -1, e^{2\pi i} = (-1)^2 = 1$$ Since we started with the general form of a complex number and simplified it to a real number (namely, $\rho$), the proof can claim that only real numbers exist and complex numbers do not. My suspicion is that the error occurs in step $4$ to $5$ , but I am not sure if that really is the case.

Answer 1

The error lies in assuming that $(\forall a,b\in\mathbb{C}):e^{ab}=(e^a)^b$.

It's worse than wrong; it doesn't make sense. The reason why it doesn't make sense is because $e^a$ can be an arbitrary complex number (except that it can't be $0$). And what is $z^w$, where $z,w\in\mathbb C$? A reasonable definition is that it means $e^{w\log z}$, where $\log z$ is a logarithm of $z$. Problem: every non-zero complex number has infinitely many logarithms: if a number $\omega$ is a logarithm, then every number of the form $\omega+2k\pi i$ ($k\in\mathbb Z$) is also a logarithm.

Answer 2

Because for complex numbers, $e^{zc}\ne(e^z)^c$ for $z,c \in \mathbb{C}$.