Clearly, $\prod_{n=1}^{\infty} (1 - z^n)$ converges to $0$ when $1 - z^{n_0} = 0$ for some $n_0$.
Suppose $z$ is a complex number with $|z| = 1$ and $1 - z^{n} \ne 0$ for all $n$.
I think $\prod_{n=1}^{\infty} (1 - z^n)$ diverges, but it is not clear how to prove this.
I know that $\prod_{n=1}^{\infty} (1 - z^n)$ converges to a non-zero number iff $\sum_{n=1}^{\infty} -z^n$ converges absolutely. Hence, if $\prod_{n=1}^{\infty} (1 - z^n)$ converges, it must converge to $0$. It is not clear how to proceed from here.
I would appreciate any hints/pointers for proving (or disproving) that $\prod_{n=1}^{\infty} (1 - z^n)$ is divergent.
If the product converges to a non-zero number then $1-z^{n } \to 1$ so $z^{n} \to 0$ as $ n \to \infty$. But $|z^{n}|=1$ so the product does not converge (to a non-zero number) for any $z$ on the unit circle.
[Convergence of an infinite product to $0$ is considered as divergence].