Define $\text{d}(\sigma)$ and $\text{maj}(\sigma)$ to be the number of descents and the major index of the permutation $\sigma$, respectively. Define a $q$-Analogue of the Eulerian Polynomials by $A_n(x, q)=\displaystyle\sum_{\sigma\in S_n} x^{\text{d}(\sigma)+1}q^{\text{maj}(\sigma)}$. I want to show that $$\sum_r [r]_q^n x^r=\frac{A_n(x, q)}{(1-x)(1-qx)\cdots (1-q^nx)},$$
where $[r]_q$ denotes $1+q+\cdots+q^{r-1}$, the $q$-analogue of the natural number $r$. I have seen a proof of this for normal powers and Eulerian polynomials, but it involves differentiation and equating coefficients to get a recurrence that is proved. However, I can't find a way to extend this to the $q$-analogue; I tried $q$-differentiation, but when I differentiate $\prod_{i=0}^n (1-q^ix)$, I don't seen to be getting $-\dfrac{[n+1]_q}{\prod_{i=0}^{n+1} (1-xq^i)}$, which is what I would expect based on the version when $q=1$. Is there an alternate way to prove the $q=1$ case that can be adapted for the $q$-analogue?