Thank you for your comments. I have found these results, which I hope may be useful.
The power of power property $(a^b)^c=a^{bc}$ is valid when:
- $a\in \mathbb{R}^+ \; \; ; \; \; b,c\in\mathbb{R}$
- For complex numbers the logarithm is multi-valued, so we must use the correct definition of power for complex numbers, which is
$$z^c=e^{c \text{Ln}(z)} \; \; ; \; \; z,c\in\mathbb{C}$$
where
$$\ln (z)=\text{Ln}(z) + 2\pi i n\; \; ; \; n\in \mathbb{Z}$$
and $\text{Ln}(z)$ is the principal value of $\ln(z)$
Then,
$$(a^b)^c=(e^{b\ln a})^c=e^{c\ln(e^{b\ln a})}=e^{c(b\ln a+2\pi i k)}=e^{cb\ln a}e^{2\pi i c k}=a^{bc}e^{2\pi i c k}\; \; ,$$
where $k\in \mathbb{z}$.
So the conclusion is that the property $(a^b)^c=a^{bc}$ is valid for $c k \in \mathbb{Z}$, then $c$ must be integer.
Finally, applying this property to our problem: $a=e$ ; $b=i \pi$ ; $c=x/\pi$; then, the third equality is valid for $x/\pi=k$; where $k\in\mathbb{Z}$. We obtain $x=k\pi$, which is consistent with $\sin(x)=\sin(k\pi)=0$.
