Let $p$ a prime number and $f(z)=\prod_{n\ge0}\left(1-z^{p^n}\right)$. One sees easily that $f$ is defined on $\left\{z\in\mathbb C_p\mid v_p(z)>0\right\}$. But can one continue it analytically on a larger disk? All I can say is that one has $$f(z^p)=\prod_{n\ge0}\left(1-z^{p^{n+1}}\right)=\frac{f(z)}{1-z^p}\qquad(1).$$ I do not know if it is enough. And if it is enough how to do? For example if one can continue the function by the relation (1), how to determine its poles and its zeroes?
Thanks in advance.
Maybe I’m missing something, but it seems to me that $f(z)\in\Bbb Z_p[[z]]$ and thus converges and is analytic in $\{z:v_p(z)>0\}$. No zeroes there, of course.
EDIT: In response to your modification to the “open” disk $\{z:v_p(z)>0\}$, let’s look at the case $p=3$, $z=5$. Now, $\lim_n5^{3^n}=-1$, so that the individual factors $1-5^{3^n}$ are approaching $1-(-1)=2$, and you’re multiplying together factors that are looking more and more like $2$. Not convergent. On the other hand, if $z\equiv1\pmod3$, you get $z^{3^n}\rightarrow1$, so $1-z^{3^n}\rightarrow0$, so that the function descdribed by your formula is well-defined, and constant zero on $\{z:v_p(z-1)>0\}$.