Proving $\prod_{n=0}^{\infty}\left(1+\frac{x}{a^n}\right)=\sum_{n=0}^{\infty}\frac{(ax)^n}{\prod_{k=1}^{n}(a^k-1)}$

Question

By trying to prove that Riemann's Zeta function is analytically expendable to the whole plane with one pole, I went aside and noticed this identity about formal power series (which are obviously everywhere convergent for complex $x$):
$$
\prod_{n=0}^{\infty}\left(1+\frac{x}{a^n}\right)=\sum_{n=0}^{\infty}\frac{(ax)^n}{\prod_{k=1}^{n}(a^k-1)},\quad\quad\forall a\in\mathbb{R}:a>1.
$$
I checked it for the first several terms on my computer and it matches. Can you help me prove this identity?
In my proof I would plug in $a=4$ and $a=3$ to prove that the Zeta function has the domain it has, although my first example to check was $a=2$. My idea was to continue Ramanujan's idea of transforming Riemann's Zeta function into Dirichlet's Eta function to represent Zeta as a quotient of two Dirichlet series that converge everywhere, denominator would be the function in the identity, with $x=2^{-s}$. That's the idea, the proof is not finished, so I am aware that this idea will probably fail in this shape. I will have to work more on it.

Answer 1

For $|a|>1$ let $$f(x)=\prod_{n=0}^\infty (1+x a^{-n}) =\sum_{m=0}^\infty c_m x^m$$ $$\sum_{m=0}^\infty c_m a^m x^m=f(ax)= (1+x a) f(x)=1+\sum_{m=1}^\infty (c_m+a c_{m-1})x^m$$ Equating the coefficients we find that $$c_0=1,\qquad c_m a^m=c_m +a c_{m-1}$$ ie. $$c_m = \frac{a}{a^m-1}c_{m-1} = \frac{a^m}{\prod_{k=1}^m (a^k-1)}$$