Prove that all the points in $D=\left\{ z \in \mathbb{C} : \mid z\mid=1 \right\}$ are singularities of the function $$ f(z)=\sum_{n=0}^{\infty} \frac{z^{n!}}{n!} $$
This was easy for the function $g(z)=\sum_{n=0}^{\infty} z^{n!}$, since I used that for $z \in E=\left\{ e^{2\pi i \cdot a/b} : a,b \in \mathbb{Z} \right\}$, $g(z)$ diverges, and then the assumption follows because $\overline{E}=D$. But I can not use that with $\ f$ because of the $1/n!$ factor. How can I improve the same argument to make it work with $\ f$?
Note that $1+ zf'(z) = g(z)$, so $f$ and $g$ have the same singularities.