Let $d$ be an integer greater than $1$ and $f(z)=\sum_{n\ge0}z^{d^n}$ be the Fredholm series. It is well-knowm that Roth-Ridout theorem imply that $f(r)$ is a transcendental number for all non zero rational $r$. By Mahler method, it is even true for any nonzero algebraic $r$. Can one extend the arguments of the proof based on Roth-Ridout theorem to recover Mahler's result?

Thanks in advance.

Let $d$ be an integer greater than $1$ and let $f(z)=\sum_{n\ge0}z^{d^n}$ be the Fredholm series. It is well-known that the Roth-Ridout theorem implies that $f(r)$ is a transcendental number for all non-zero rational $r$. By Mahler’s method, it is even true for any nonzero algebraic $r$. Can one extend the arguments of the proof based on the Roth-Ridout theorem to recover Mahler's result?

Thank you in advance.