I would like to show $\sum_{j=0}^n (-1)^j {n \brack j}_q =0$ for n odd, or preferably even more generally that $\sum_{j=0}^n (-1)^j {n \brack j }_q =\frac{1}{2}((-1)^n+1)(q;q)_{\frac{n}{2}}$.
Using the q-binomial theorem I've managed the following:
$\sum_{j=0}^n (-1)^j {n \brack j}_q = \sum_{j\geq0} \frac{(q^{n-j+1};q)_n}{q;q)_n}(-1)^j = \frac{((-q)^{n-j+1};q)_{\infty}}{(-1;q)_{\infty}}$
Where I have ${ n \brack j}_q=\frac{(q;q)_n}{(q;q)_j(q;q)_{n-j}}$ and $(q;q)_n =\prod_{i=0}^n(1-q^{i+1})$.
How do I go from there to my identity?