Tried to prove that $e$ is irrational but ended up proving it's not a real number. Assume $e=p/q$, with integers $p,q$. Then we have:
\begin{align}
&qe=p\\
&(qe)^{2i\pi}=p^{2i\pi} , \text{ since $e^{2i\pi}=1$:} \\
&q^{2i\pi}=p^{2i\pi}, \text{since the exponents are the same:} \\
&q=p
\end{align}
that would imply $e=p/q=1$. But nowhere in the "proof" uses the fact that $p,q$ are integers. If we take $p,q$ complex $e$ might not even be a number.
Where is the mistake?
 A: You went wrong in the last step. As you're fully aware, $e^{2i \pi} = 1$, just as $e^0 = 1$. This means that the complex exponential function is no longer injective, literally meaning that $e^z = e^w$ no longer implies $z = w$, with $z = 2 i \pi$ and $w = 0$ being a counterexample.
So, what does $e^z = e^w$ actually imply? We have
$$e^z = e^w \implies \frac{e^z}{e^w} = 1 \implies e^{z - w} = 1.$$
If we write $z - w = x + iy$, where $x, y \in \Bbb{R}$, this yields
$$1 = e^{x + iy} = e^x(\cos(y) + i \sin(y)),$$
which implies $e^x = 1$ from taking the modulus of both sides, and $y = 2\pi k$ for some integer $k$. Since $x \in \Bbb{R}$, we must have $x = 0$, so
$$z - w = 2 i \pi k$$
for some $k \in \Bbb{Z}$.
So, given $p^{2 i \pi} = q^{2 i \pi}$, we can conclude, by definition of complex exponentiation,
$$e^{2 i \pi \log p} = e^{2 i \pi \log q}$$
and thus, for some $k \in \Bbb{Z}$,
$$2 i \pi \log p = 2 i \pi \log q + 2 i \pi k \implies \log(p / q) = k \implies p / q = e^k.$$
In this case, $k = 1$; there is no contradiction.
A: $$q^{2\pi i}=p^{2\pi i}\to e^{i\cdot 2\pi \ln q}=e^{i \cdot 2\pi \ln p}\tag{1}$$
Remember that $e^{i\theta}=\cos\theta +i\sin \theta$:
Because of the periodicity of $\cos$ and $\sin$, the only conclusion you can draw from $(1)$ is:
$$2\pi\ln(q)=2\pi(\ln(p)+k) : k\in \Bbb Z$$
Which we may simplify to: $$\ln(q)-\ln(p)=k\implies \frac qp = e^k$$
Here, $k=-1$ trivially.
