$\DeclareMathOperator\maj{maj}\DeclareMathOperator\inv{inv}$Let $t_{a,b}$ be the numbers
$$
 t_{a,b} \mathrel{:=} \lvert\{ \pi \in S_{a+b} : \maj(\pi)=a \text{ and }  \maj(\pi^{-1})=b \}\rvert.
$$
Here, $S_{a+b}$ denotes the set of permutations of $1,2,\dotsc,a+b$.
By a result of Foata, one can also look at the pair of statistics $(\maj, \inv)$, and a few other combinations — these pairs of statistics will produce the same numbers.


Now, according to the OEIS entry [A090806][1], it is proved by Garsia–Gessel that
\begin{equation}
\label{*}
\tag{*}
 \sum_{a,b} t_{a,b} q^a t^b = \prod_{i,j \geq 1} \frac{1}{1-q^i t^j}.
\end{equation}
I cannot see exactly where in their paper one can deduce this.

**My attempt:**
I have tried to prove this myself (mainly by resorting to RSK, the Cauchy identity,
and some symmetric function identities).
This leads to the following (which appears in Stanley's EC2):
\begin{equation}
 \sum_{n \geq 0} \frac{z^n}{(1-q)^n[n]_q!(1-t)^n [n]_t!} \sum_{\pi \in S_n} t^{\maj(\pi)} q^{\maj(\pi^{-1})} 
 =
 \prod_{i,j \geq 0} \frac{1}{1-z q^i t^j},
\end{equation}
where $[n]_q! \mathrel{:=} [1]_q [2]_q \dotsm [n]_q$, and $[n]_q = 1+q+q^2+\dotsb + q^{n-1}$.
However, I do not see some short way to deduce the above generating function from this.

**Question:** Is there some alternative (more recent?) reference where \eqref{*} is 
stated and easily referenced? Alternatively, someone who can see exactly where in the paper \eqref{*} is proven?



  [1]: https://oeis.org/A090806

<cite authors="Garsia, A. M.; Gessel, I.">_Garsia, A. M.; Gessel, I._, [**Permutation statistics and partitions**](http://dx.doi.org/10.1016/0001-8708(79)90046-X), Adv. Math. 31, 288-305 (1979). [ZBL0431.05007](https://zbmath.org/?q=an:0431.05007).</cite>