I have seen a proof of $\pi$ being transcendental by conclude that transcendental number powered by algebraic number must be transcendental and algebraic number powered by algebraic number must be algebraic.
So, from the equation $e^{i\pi} = -1$ and $i$ is algebraic so $\pi$ must be transcendental.
But then why is $e^{\pi}$ transcendental?
You can use the Gelfond-Schneider theorem to prove that $e^\pi$ is transcendental. The theorem states that:
If $a$ and $b$ are algebraic numbers with $a ≠ 0,1$ and $b$ non-rational, then any value of $a^b$ is a transcendental number.
Using the equality:
$$ e^\pi = \left( e^{i \pi} \right)^{-i} = (-1)^{-i}, $$
we obtain $a=-1$ and $b=-i$, the latter of which is algebraic and non-rational. Therefore $e^\pi$ is transcendental.