We know that if $x[n]$ has Z-transform, and an RoC of $\alpha < |z|<\beta$ then the z transform of $p^n x[n]$ is $X(\frac{z}{p})$ with an RoC of $ p\alpha < |z| < p\beta$.
But I've seen another theorem regarding Z-transform of X, that if and only if ${r_0}^{-n} x[n]$ is absolutely summable, then $r_0$ is in the RoC of $X(z)$. I cannot prove this using the aforementiond property of Z-transform, and I'm curious how can I prove this.
Using the first property, the RoC of the new signal should be: $ \frac{\alpha}{p} < |z| < \frac{\beta}{p}$, which does not necessarily include $r_0$.
So after a day of thinking about this, I finally came up with an answer.
We know that the z-transform of a signal is defined as $$\sum_{n=-\infty}^{+\infty}{x[n]z^{-n}}\quad \quad RoC: \alpha < |z| < \beta$$
And the Z-transform of the signal exists for all values of $|z|$ that lie inside the RoC. So the Z-transform of the signal for an specific value, $r_0$, that lies in the RoC will be: $$X(r_0)=\sum_{n=-\infty}^{+\infty}x[n]{r_0}^{-n}$$
Which is the exact same signal in the title of the question. Hence there are no direct relations to the properties of the Z-transform, but the definition of Z-transform itself. Because the z-transform only exists if and only if the signal is abosolutely summable in a given RoC, each point that lies inside the RoC will as well follow this property.