We define the q-analog of the Eulerian polynomial $A_n(x)$ as $A_n(x,q) = \sum_{\sigma \in S_n} x^{d(\sigma)+1}q^{maj(\sigma)}$, where $S_n$ is the set of permutations of $n$ elements, $d(\sigma)$ is the cardinality of the descent set of the the permutation $\sigma$ and $maj(\sigma)$ is the major index of $\sigma$
I need to show that with this definition we have $$\sum_r[r]^{n}_qx^{r}=\frac{A_n(x,q)}{(1-x)(1-qx)...(1-q^nx)}$$
I have been stuck in this problem for a long time. Could anyone give me a hint on how to approach it?