Where is the error in this proof for showing that $f:\mathbb{R}\rightarrow\mathbb{R},x\mapsto\sin{(2\pi x)}$ is constant? Show that $f:\mathbb{R}\rightarrow\mathbb{R},x\mapsto\sin{(2\pi x)}$ is constant.

$$1=1^x=(e^{2\pi i})^x=e^{2\pi ix}=\cos{(2\pi x)}+i\sin{(2\pi x)}$$
Where is the error in this "proof"? Does it not hold for all $x\in\mathbb{R}$, if yes which ones?
 A: The mistake is
$$
\left(e^{2 \pi i }\right)^{x}=e^{2 i \pi x}
$$
holds only for $x \in \mathbb{Z}$.
A: Notice that $(-2)^2 = 4$.  So $[(-2)^2]^{\frac 12} = 4^{\frac 12} = 2$.
But $[(-2)^2]^{\frac 12} = (-2)^{2\frac 12} = (-2)^1 = -2 \ne 2$.
What went wrong?
The assumption that $(a^m)^n = a^{mn}$ only holds if either $m, n \in \mathbb Z$  or if $a > 0; a \in \mathbb R; m,n \in \mathbb R$. 
If we define $a^m = a*.... *a; m$ times then $(a^m)^n = a^{mn}$ follows by associativity.
If we define $b^{\frac mn} = \sqrt[n]{b^m}; b > 0; \frac mn \in \mathbb Q$ then $(b^q)^r = b^{qr}$ follows,  by putting the fractions under a common denominator. This reduces  to a radical base and integer powers.
If we define $b^x = \lim_{q\to x} b^q; b > 0$ then $(b^x)^y = b^{xy}$ can be deduced  by the continuity of limits.
But when we define a fundimentally different definite for complex powers as $e^{iy} = \cos y + i \sin y$ then $(e^{iy})^z = (\cos y + i \sin y)^z \ne \cos yz + i \sin zy= e^{iyz}$ just plain doesn't follow. 
