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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?

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