We all know that $\sum_{i=0}^{n}{n \choose i}=2^{n}$. Is there a similar result regarding the qbinomial coefficients? (a.k.a Gaussian binomial coefficients)  $\sum_{i=0}^{n}{n \choose i}_{q}=?$
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The identity $\prod_{i=0}^{n1} (1+xq^i) = \sum_{k=0}^n x^k q^{{k\choose 2}}{n\choose k}_q$ is the $q$binomial theorem. A combinatorial proof based on integer partitions is mentioned on page 68 of Enumerative Combinatorics, vol. 1, 2nd ed. There is also given a combinatorial proof based on finite fields. For the online version at http://math.mit.edu/~rstan/ec/ec1.pdf, see pages 7475. 


There are many possibilities, e.g. $\sum_{i=0}^{n}q^i{n \choose i}_{q^2}=(1+q)(1+q^2)...(1+q^n)$ or $\sum_{i=0}^{n}q^{i(i+1)/2}{n \choose i}_{q }=(1+q)(1+q^2)...(1+q^n).$ 

