TITLE: A "quantum" identity: in search of a proof -Part II
QUESTION [12 upvotes]: As usual, denote $[n]_q=1+q+\cdots+q^{n-1}=\frac{\,\,1-q^n}{1-q}$ and $[n]_q!=[1]_q[2]_q\cdots[n]_q$. Furthermore, we write
$$\binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!}.$$
As a follow up on this MO question, I propose a $q$-analogue identity.

Question. Can you show that
  $$\sum_{k=0}^nq^{(y-n+1)k}\binom{x+k}k_q\binom{y-k}{n-k}_q
=\sum_{k=0}^nq^{n-k}\binom{x+y-k}{n-k}_q\,\,\,?$$

It would be great if we can see alternative proofs? I've a bias for combinatorial arguments. :-)

REPLY [7 votes]: Actually this is a partial case of $q$-Vandermonde identity.
In the notation $$[a;b]_q^n:=(a-b)(a-qb)\dots (a-q^{n-1}b)=a^n(b/a;q)_n,$$ as in this terminology question, we have 
$$
[a;c]_q^n=\sum {\binom{n}{k}}_q[a;b]_q^{n-k}\,[b;c]_q^{k},
$$
which is an equivalent form of $q$-Vandermonde identity.
Substituting $a=q^{n-y-1}$, $b=1$, $c=q^{x+1}$ we get $$\sum_{k=0}^nq^{(y-n+1)k}\binom{x+k}k_q\binom{y-k}{n-k}_q=\binom{x+y+1}n_q,$$
replacing $x,y$ to $x+y-n,n$ (and $k$ to $n-k$ in the sum) we get that RHS also equals $\binom{x+y+1}n_q$.