Can this sum over the q-Pochhammer symbol be simplified?

While considering the problem of the expected value of a dice fixing strategy on a two-sided die that comes up as $1$ with a probability of $\alpha$ and $0$ otherwise. I was studying the strategy where one fixes every die if they are all $1$'s and otherwise fixes only one die. I found that the expected value of this strategy satisfies the functional equation $$A(x)=C(x)-\frac{\alpha x}{1-x} A(\alpha x)$$ where $C$ is some particular rational function (although, I'd be interested in solutions for any or every rational $C$). By making infinitely many substitutions of this equation into itself, we can solve this as: $$A(x)=\sum_{n=0}^{\infty}\left(\prod_{k=0}^{n-1}\frac{-\alpha^{k+1} x}{1-\alpha^kx}\right)C(\alpha^n x)=\sum_{n=0}^{\infty}\frac{(-1)^n\alpha^{n(n+1)/2}x^nC(\alpha^nx)}{(x;\alpha)_n}$$ where $(x;\alpha)_n$ is the q-Pochhammer symbol defined as $$(x;\alpha)_n = \prod_{k=0}^{n-1}(1-x\alpha^k).$$ Is there a simpler form for the sum for $A$?

I am hopeful that there is such a form since I the Wikipedia page on q-Pochhammer symbols lists the following identity: $$(x;q)_{\infty}=\sum_{n=0}^{\infty}\frac{(-1)^nq^{n(n-1)/2}x^n}{(q;q)_n}$$ which looks very similar to what I'm trying to get. In fact, this lets us calculate particular values of our generating function for certain $C$. For instance, if $C(x)=\frac{1}{1-x}$, we would find that at $x=\alpha$, we would have $$A(\alpha)=\sum_{n=0}^{\infty}\frac{(-1)^n\alpha^{n(n+1)/2}\alpha^n}{(\alpha;\alpha)_{n+1}}=\frac{1-(\alpha;\alpha)_{\infty}}{\alpha}.$$ However, I don't see any other useful identites.