I would like to find the domain of convergence of $\displaystyle{\prod_{n=1}^\infty \left(1+\left(1+\frac{1}{n}\right)^{n^2}z^n\right) }$, where $z$ is a complex number. From my understanding, an infinite product of complex numbers $\displaystyle{\prod_{n=1}^{\infty} 1+ a_n}$ converges iff {$a_n$}$_{n=1}^\infty$ converges to $0$ and if $q\in \mathbb{N}$ is chosen such that for every $n\geq q$, $1+a_n \in G_0=\mathbb{C}\setminus (-\infty,0]$, then $\displaystyle{\sum_{n=q}^\infty \log(1+a_n)} $ converges, where $\log$ is the principle branch of the log.
I believe that the domain of convergence is {$z\in \mathbb{C}: |z|<\frac{1}{e}$}, as when I look at the limit of the $a_n$, it converges to zero when $|z|<\frac{1}{e}$. Now I would like to fix $z$ with $|z|<\frac{1}{e}$. Since the $a_n$ converge to 0, there is some $q$ sufficiently large such that $|a_n|<1$ thus for all $n\geq q$, $1+a_n\in G_0$. Thus I believe it remains to show that the sum of the logs converges. My initial idea was to look at the power series expansion of the principle branch of the log on $|z-1|<1$:
$\displaystyle{\sum_{n=q}^\infty \log(1+a_n)}=\sum_{n=q}^\infty \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)}((1+\frac{1}{n})^{n^2}z^n)^{k+1} $
My guess is that this should converge for fixed $z$ with $|z|<\frac{1}{e}$, but I'm not sure how to proceed with showing that this series converges.
Very good work so far! All you need to finish it off is some inequality like $|\log(1+w)| \le 2|w|$ for $|w|\le\frac12$ (the exact constants aren't important, as long as the $\frac12$ is strictly less than $1$ to avoid the singularity of $\log(1+w)$ at $w=-1$) and the comparison test. Inequalities like this, of functions analytic on a closed disk, exist because of the maximum modulus principle applied to $\dfrac{\log(1+w)}w$.