In my complex variable course we are studying series convergence, using comparison test I determined that the series
\begin{equation} \sum_{n=0}^{\infty}\frac{z^{n!}}{n!} \end{equation}
has a convergence radius $R\leq 1$, so we know that at some point past the unitary disk is the first singularity, actually my professor told us that this series has a natural boundary, how could I prove it? he also asked for a single singularity in the boundary (as an example), any hint?
The series $$f(z)=\sum_{n=0}^\infty \frac{z^{n!}}{n!}$$ has a convergence radius $R=1$ and analytic in the unit disc $\mathbb{D}=\{|z|<1\}.$ We can show that the derivative $$f^\prime(z)=\sum_{n=0}^\infty z^{n!-1}$$ has $\partial\mathbb{D}$ as a natural boundary directly (by showing the function tends to infinity on every radius whose argument is a rational multiple of $\pi$, see Conrad's comment) or by gap theorem.
Suppose that $f(z)$ has its analytic continuation on $\mathbb{D}\cup \{|z-z_0|<\delta\}$ with $z_0\in \partial\mathbb{D}$. Then $f^\prime(z)$ should have its analytic continuation on $\mathbb{D}\cup \{|z-z_0|<\delta\}$. But it is impossible as stated above. So $f(z)$ has $\partial\mathbb{D}$ as a natural boundary.