lacunary functions wander, part one

Consider for a moment:
$\prod_{n=1}^{\infty} \sqrt[2^{n}]{1+\frac{1}{2^{n}}}$
One fruitful way to generalize this is to consider a function of $|q|<1$ like so:

$(q)=\prod_{n=1}^{\infty} (1+q^{n})^{q^{n}}$

$\log$$(q) = \sum_{n=1}^{\infty} q^{n} \log(1+q^{n})$

$\sum_{n=1}^{\infty} q^{n} \log(1+q^{n}) = \sum_{n=1}^{\infty} q^{n} \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m}q^{mn}$

$\sum_{n=1}^{\infty} q^{n} \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m}q^{mn} = \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m} \sum_{n=1}^{\infty} q^{mn} q^{n} = \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m} \sum_{n=1}^{\infty} q^{n(m+1)}$

$\sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m} \sum_{n=1}^{\infty} q^{n(m+1)} = \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m} \left[\frac{1}{1-q^{m+1}}-1\right]$

Which is just
$\sum_{m=1}^{\infty} \left[\frac{(-1)^{m+1}}{m(1-q^{m+1})}\right] - \log(2)$

Therefore:

$(q)=\prod_{n=1}^{\infty} (1+q^{n})^{q^{n}} = \frac{1}{2}\exp\left(\sum_{m=1}^{\infty} \left[\frac{(-1)^{m+1}}{m(1-q^{m+1})}\right]\right)$

Here is a plot of अ$(q)$ on the unit disk. There are some artifacts, but nothing major. Enjoy: