This proof tries to show that
$ \mathbb{R} = \mathbb{C} $
$Let z \in \mathbb{C}, \exists r \in \mathbb{R}^{+}, \theta \in [0,2\pi]:z=re^{i\theta}$ $\\z=re^{i\theta\frac{2\pi}{2\pi}} = r(e^{2\pi i})^{\frac{\theta}{2\pi}}= r(1)^{\frac{\theta}{2\pi}}=r \\so z \in \mathbb{R} \\\rightarrow\mathbb{C} \subset \mathbb{R} \\ therefore \mathbb{C}=\mathbb{R}$
This proof is clearly false but I can't seem to find any algebraic mistake in it.
The equality $$(a^b)^c = a^{bc}$$ is not true for complex numbers (it is only true in general if $a$ is a positive real, and $b,c$ are real), so your proof is false.