Product of one minus geometric progression Consider the geometric sequence $a_n:=ar^{n-1}$, $n=0,1,\ldots$. 
We know that the product
$\Pi_{i=0}^n a_i=(\sqrt{a_1\cdot a_{n+1}})^{n+1}$. 
Is there a formula for $\Pi_{i=0}^n (1-a_i)$?
 A: In a standard $q$-notation, the product of your interest is
$$
(z;q) _ n:=\prod_{j=0}^{n-1}(1-zq^j).
$$
The formula known as the $q$-binomial theorem expresses this product as sum:
$$
(z;q) _ n=\sum _{k=0}^n {\genfrac{[}{]}{0pt}{}{m}{n}} _q (-z)^kq^{k(k-1)/2},
$$
where the $q$-binomial coefficients are defined, e.g., in this question.
A: If $a = -1$ you get $\displaystyle \left(1 + \frac{1}{r}\right)\cdot 0 \cdot \prod_{i=k}^n (1 + r^k)= 0$.  The last factor is a generating function (free book inside!)   counting the number of partitions of $k$ into distinct parts with parts at length at most $n$.  
Let's look at the partitions of $5$.
\[ (5) \quad (4+1)\quad (3+2)\quad (3+1+1)\quad (2+2+1)\quad (2+1+1+1)\quad (1+1+1+1+1)\] 
This would appear as the $r^5$ term in that series.  However, many of these do not have distinct parts. So we rule out the last four.
\[ (5) \quad (4+1)\quad (3+2)\]
However, with your two additional factors, you're allowing upto one $0$ and one $-1$ in your partition.
\[ (5) \quad (4+1)\quad (3+2)\quad(6+(-1))\quad (5+0) \quad (4+1+0)\quad (3+2+0)\quad(6+0+(-1))\dots\]

Your sequence begins $\frac{a}{r}, a, ar, ar^2, ar^3, \dots $.  The coefficient of $\displaystyle \prod_{i=0}^n (1 + br^{i-1})$ gives a weighted sum of partitions into distinct parts.  Let's show how that count works. (NOTE: my $b$ is your $-a$).


*

*5 has weight $b$

*(4 + 1) and (3+2) have weight $b^2$

*(3+2+0) has weight $b^3$
and so on. 


In light of this, I doubt there is a closed formula for this infinite product beyond this or similar interpretations as counting partitions, (see Wadim's answer) but these are extremely interesting functions!
Two guys who study partitions for a living are Herbert Wilf and George Andrews
