Okay, I think that I know what you mean. I tried to calculate it, but there is some small error in it. But I think it is worth to share. There is missing some sign, so your homework is to find the missing sign. Here is the solution:
$ \begin{align*} \prod^{n}_{k=1} (1 + yq^k) &= \frac{(-yq;q)_{\infty}}{(-yq^{n+1};q)_{\infty}} \\ &= {}_1 \phi_0(q^{-n}; q, -yq^{n+1}) \\ &= \sum^{\infty}_{m=0} \frac{(q^{-n};q)_m}{(q;q)_m} (-yq^{n+1})^m \\ &= \sum^{n}_{m=0} y^m (-1)^m q^{m(n+1)} \frac{(1-q^{-n})...(1-q^{-n+m-1})}{(1-q)...(1-q^m)} \\ &= \sum^{n}_{m=0} y^m (-1)^m \frac{(q^{n+1}-q)...(q^{n+1}-q^m)}{(1-q)...(1-q^m)} \\ &= \sum^{n}_{m=0} y^m q \cdot q^2 \cdot ... \cdot q^m (-1)^{2m} \prod^{m-1}_{k=0} \frac{(1-q^{n-k})}{(1-q^{k+1})} \\ &= \sum^{n}_{m=0} y^m q^{m(m+1)/2} \frac{(q)_n}{(q)_m(q)_{n-m}} \end{align*} $
TADA! :D