The q-binomial theorem states that $$\sum_{n=0}^{\infty}\frac{\left(a;q\right)_n}{\left(q;q\right)_n}z^n =\frac{\left(az;q\right)_\infty}{\left(z;q\right)_\infty}$$. Is there a similar closed-form expression for when the top index now runs over the negative integers? That is,
$$\sum_{n=0}^{\infty}\frac{\left(a;q\right)_{-n}}{\left(q;q\right)_n}z^n = ?$$
Basic rewriting of the top term did not help.
After rewriting, I get a form that is quite similar to the below, which I'm also interested in finding a closed-form expression for
$$\sum_{n=0}^{\infty}\frac{\left(\frac{\left(1-\lambda\right)\left(1-q\right)}{\lambda q}\right)^{n+1}}{(q;q)_n(\frac{1}{\lambda};\frac{1}{q})_{n+1}} = ? $$