irrational numbers - Proof for the irrationality of $eπ$ ('complex'ly?).*

Some of you may have seen the video:

https://youtu.be/DLWpj34UNRk

It is a proof of the irrationality of $eπ$ by Ron (14years) , presented by 'blackpenredpen' Youtuber.

The proof goes something like this:

We know that $e^{iπ}=-1$ ("Euler's famous identity").

$e^{iπ}=i^2$

So, $π=\frac{2\ln{i}}{i}$

We must prove that $eπ$ is irrational.

Let us assume, on the contrary, that $eπ$ is rational.

$eπ=\frac{a}{b}$ where $a$ and $b$ are rational numbers.
$$eπ=e\frac{2\ln{i}}{i}=\frac{a}{b}$$
$$2eb\ln{i}=ai$$
$$\ln{i^{2eb}}=ai$$

$$\ln{(-1)^{eb}}=ai$$
$$(-1)^{eb}=e^{ai}$$
Squaring on both sides,
$$1^{eb}=e^{2ai}$$
$$1=e^{2ai}$$
$$e^0=e^{2ai}$$
When the bases are identical, the powers are equal.
$$0=2ai$$
$$a=0$$
Hence,

$$eπ=\frac{0}{b}$$
$$eπ=0$$
This is not possible and therefore gives us a contradiction.

This contradiction has arisen because of our incorrect assumption the $eπ$ is rational.

We conclude that $eπ$ is irrational.

(*) Is this proof legitimate? Or is there something dubious about it? If so, please point it out.