For the $q$-exponential $$e_q(u) = \sum_{n=0}^{\infty} \frac{u^n}{[n]_q!}$$ with $[k]_q=\frac{1-q^k}{1-q}$ and $[n]_q! = [n]_q [n-1]_q \cdots [1]_q$, we don't have the property $e_q(u) e_q(v) = e_q (u+v)$. I learned this the hard way when trying to compute the infinite product $$\prod_{i=0}^{\infty} e_q (x q^i)$$ where $x$ is an indeterminate.

- Does anyone see a way to get a closed form expression for this product?

Any help would be greatly appreciated!