I have already proven this identity:
$\prod_i (1+st^i) = 1 + \sum_r \frac{s^rt^{r(r+1)/2}}{(1-t)(1-t^2)\cdots(1-t^r)}$
I expanded the product, grouped the s terms, and then made an argument about the number of integer partitions with r parts being represented by the series $t^{r(r+1)/2}/(1-t)(1-t^2)\cdots(1-t^r)$.
Now I must prove this identity:
$\sum_k \frac{(1+a)(1+at)\cdots(1+at^{k-1})z^kt^k}{(1-t)\cdots(1-t^k)} = \prod_i \frac{1+azt^i}{1-zt^i}$
I'm trying to do so by applying the first identity. When I do that the sum becomes
$\sum_r \left[\frac{z^r}{(1-t)\cdots(1-t^r)} a^rt^{r(r+1)/2} \prod(1+zt^k+z^2t^{2k}+\cdots)\right]$
The part in the fraction on the left matches where I want to be in the solution but I can't seem to find any manipulations that give me the rest of it. That product doesn't depend on $r$ either which makes me think that maybe I'm barking up the wrong tree. So now I'm looking for some ideas.
Your proven identity looks like the q-binomial theorem, and the terms can be rewritten with q-shifted factorial. I don't have the answer for your question, but it might exist some useful identities in q-series that can help to prove the identity. If you have already use this, then ignore my answer.