Consider the expression
$$G(x,a) = \frac{1}{((1-a)x;a)_{\infty}}$$
Based on:
Infinite sum involving ascending powers
It follows that in the limit as $a \rightarrow 1$
$$\frac{1}{((1-a)x;a)_{\infty}} = \sum_{i=0}^{\infty} \left[ \frac{1}{i!}x^i\right] = e^x $$
What I'm curious is. How do I express
$$ G(u+v,a) \ \& \ G(uv,a)$$
In terms of
$$ G(u,a) , G(v,a),u,v$$
For example
$$e^{xy} = (e^{x})^y $$ $$e^{x+y} = e^x e^y $$
If $x,y \in \Bbb{R}$ or $x,y \in \Bbb{Q}$ whereas for the latter case we treat the exponential as a multivalued function.
Can these types of identities be generalized for the function I have given?