Similar to Macmahon's theorem, why does $\sum t^{\text{maj}(w)}=\sum t^{\text{inv}(w)}$?

In 1913 Percy MacMahon proved that the distribution of the major index on all permutations of a fixed length is the same as the distributin of inversions. I'm trying to understand the identity $$\sum_{w\in S_n\cdot(1^{k_1},2^{k_2},\dots,r^{k_r})}x^{\text{maj}(w)}=\sum_{w\in S_n\cdot(1^{k_1},2^{k_2},\dots,r^{k_r})}x^{\text{inv}(w)}$$ where $k_1+\cdots+k_r=n$, and $w\in[r]^n$. With help, I've come to understand that if $w(n)=s$, then $w'=(1 2 \cdots r)^{r-s}\circ w$ has $\text{maj}(w')=\text{maj}(w)-(k_{s+1}+\cdots+k_r)$. I don't see how that leads to the result however. I've been searching around for MacMahon's theorem, but haven't found anything that actually proves this identity, and was hoping it could be resolved here. Thank you.

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It seems you are after the bijective proof due to D. Foata and M. P. Schützenberger in Major index and inversion number of permutations, Math. Nachr. 83 (1970), 143–159. –  Did Feb 17 '12 at 9:28
@Didier: I think that the 1970 paper deals only with permutations of $[n]$; Nastassja is looking at permutations of $n$-tuples from $[r]$, where $r\le n$. –  Brian M. Scott Feb 17 '12 at 10:57