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