Infinite sum involving ascending powers I was working on a problem involving finite differences and came across the following sum as a general solution to a problem I was working with
$$ 1 + x + \frac{1}{1 + a}x^2  + \frac{1}{1 + a} \frac{1}{1 + a + a^2}x^3 + \frac{1}{1 + a} \frac{1}{1 + a + a^2} \frac{1}{1 + a + a^2 + a^3}x^4 ... $$
I note that the following sum can be rewritten as 
$$ 1  + \sum_{i = 1}^{\infty} \left[ \frac{(1-a)^i}{(1-a)(1-a^2)...(1-a^i)}x^i \right]  $$
This of course appears to be intimately related to Euler's function
$$ \phi(n) = \prod_{i=0}^{\infty}\left[1 - q^i \right]$$
Except I'm considering finite cases for each component of the sum. I did a little looking around and it appears this is known as a "q-pochammer" symbol. Whereas each term is equal to
$$(1;a)_n = (1 - a)(1 - a)(1 - a^2) ... (1 - a^n)$$
Thus,my sum reduces to
$$ 1  + \sum_{i=1}^{\infty}{\frac{((1-a)x)^i}{(1;a)_i}} $$
I wish wolfram alpha had pochammer support since this looks like it should be "trivial" but I have no idea how to analyze it myself
Any help would be appreciated to find the closed form.

A finite difference approach that I can motive is to consider the function
$$E(x,i) = \frac{((1-a)x)^i}{(1;a)_i} $$
Then the function $g(x,i)$ such that 
$$g(x,i+1) - g(x,i) = E(x,i)$$
Yields a closed form to the sum by evaluating g at i = 1 and i = infinity and taking their difference. 
I can' seem to find a symbolic way of processing this however.

Actually, the standard notation is as follows.
$$(A ; q)_0 := 1, \quad (A ; q)_n := (1 - A)(1 - Aq)\cdots (1 - Aq^{n-1})\quad (n \ge 1)$$
Also, $(A; q)_\infty := \lim_{n\to \infty} (A; q)_n$. Using this, we can write your series as 
$$\sum_{n = 0}^\infty \frac{(0;a)_n}{(a;a)_n}z^n$$
where $z = (1 - a)x$. By the $q$-binomial theorem, this equals 
$$\frac{(0z;a)_\infty}{(z ; a)_\infty} = \frac{1}{((1 - a)x;a)_\infty},$$
if $|a| < 1$ and $|(1 - a)x| < 1$.

Note: If $|a| > 1$, the series can be written
$$\sum_{n = 0}^\infty \frac{(-1)^n(\frac{1}{a})^{n(n+1)/2}[(1 - a)x]^n}{\left(\frac{1}{a};\frac{1}{a}\right)_n}.$$
By the Euler transformation 

$$\sum_{n = 0}^\infty \frac{(-1)^nq^{n(n+1)/2}z^n}{(q;q)_n} = (z;q)_\infty,\quad |q| < 1$$

we have
$$\sum_{n = 0}^\infty \frac{(-1)^n(\frac{1}{a})^{n(n+1)/2}[(1 - a)x]^n}{\left(\frac{1}{a};\frac{1}{a}\right)_n} = \left((1 - a)x; \frac{1}{a}\right)_\infty$$
since $|1/a| < 1$.
