TITLE: Factoring $\frac{1}{1-rx}$ into an infinite products of polynomials
QUESTION [7 upvotes]: I am looking for examples of sequences of polynomials $(p_{k}(x))_{k=1}^{\infty}$ with positive integer coefficients where $p_{k}(0)=1$ for all $k\geq 1$ and where there is a positive integer $r$ where 
$$\frac{1}{1-rx}=\prod_{k=1}^{\infty}p_{k}(x)$$
whenever $x\in(0,\frac{1}{r})$.
One can easily construct many examples of polynomials $p_{k}(x)$ that satisfy this infinite product formula using recursion. Let us therefore limit the scope of this question to sequences $(p_{k}(x))_{k=1}^{\infty}$ that have a closed-form expression (unless there is a very good reason to deviate) and are sequences that Srinivasa Ramanujan may be interested in.
Let me get us started with a couple of examples.
Ex 1: $$\frac{1}{1-x}=\prod_{k=0}^{\infty}(1+x^{2^{k}})$$
More generally, if $a_{k}=b_{0}\dots b_{k-1}$ for all $k\geq 0$, ($a_{0}=1$ being an empty product), then
$$\frac{1}{1-x}=\prod_{k=0}^{\infty}p_{k}(x)$$
where $$p_{k}(x)=\frac{x^{a_{k+1}}-1}{x^{a_{k}}-1}=
1+x^{a_{k}}+x^{2a_{k}}+\dots+x^{a_{k+1}}.$$
Ex 2: $$\frac{1}{1-2x}=\prod_{k=1}^{k}(1+x^{k})^{a(k)}$$
where $a(k)$ is sequence A-306156 in the OEIS. Similar products exists for $\frac{1}{1-rx}$ for all $r>1$.
Polynomials that satisfy these conditions naturally arose in set theory in my investigations of rank-into-rank cardinals which float near the top of the large cardinal hierarchy.

REPLY [2 votes]: Slightly generalized from Ex. 1: 
$$ \frac{1}{1-r x} = \prod_{j=0}^\infty \sum_{i=0}^{m-1} (r x)^{i m^j} $$
for integers $m \ge 2$.