By Cauchy identity, $${}\_1\phi\_0(a;—;q,z)=\sum\_{n\geq0}\frac{(a;q)\_n}{(q;q)\_n}z^n=\frac{(az;q)\_{\infty}}{(z;q)\_\infty},\quad|z|<1,|q|<1,$$
we can obtain the $q-$analogue of $(1-z)^{-a}(1-z)^{-b}=(1-z)^{-a-b},$ $${}\_1\phi\_0(a;—;q,z){}\_1\phi\_0(b;—;q,az)={}\_1\phi\_0(ab;—;q,z),$$
which is $$\sum\_{n\geq0}\frac{(a;q)\_n}{(q;q)\_n}z^n\sum\_{m\geq0}\frac{(b;q)\_m}{(q;q)\_m}(az)^m=\sum\_{n\geq0}\frac{(ab;q)\_n}{(q;q)\_n}z^n.$$


Comparing the coefficients of $z$ in the both side of equation above, the identity in the title can be established as
$$\sum\_{i=0}^n\frac{(a;q)\_i}{(q;q)\_i}\frac{(b;q)\_{n-i}}{(q;q)\_{n-i}}a^{n-i}=\frac{(ab;q)\_n}{(q;q)\_n}.$$


However, what I really concern is that there is any possible for us to prove this identity directly without utilizing Cauchy identity. Maybe we can use **mathematical induction** on *$n$* or some other **combinatorial** or **algebraic** proofs?



How to prove that $\sum_{i=0}^n\frac{(a;q)_i}{(q;q)_i}\frac{(b;q)_{n-i}}{(q;q)_{n-i}}a^{n-i}=\frac{(ab;q)_n}{(q;q)_n}$?
Here's a proof that doesn't involve Cauchy's identity. I will begin by stating two elementary lemmas.


**Lemma 1:** Let $f \in \mathbb{C}[x,y,z]$ be a polynomial. If $f$ vanishes on the set
$$\{ (a,a^i,a^j) : a \in \mathbb{C} , i,j \in \mathbb{N}\_{>0} \},$$
then $f$ is identically 0.


**Lemma 2:** Let $\binom{n}{k}\_q :=\frac{(q;q)\_n}{(q;q)\_k (q;q)\_{n-k}}$ denote the $q$-binomial coefficients. Then the following Vandermonde-like identity holds:
$$\sum\_{i} \binom{i+A-1}{i}\_q \binom{n-i+B-1}{n-i}\_q q^{A(n-i)} = \binom{n+A+B-1}{n}\_q.$$




---


Given these two lemmas, we proceed as follows: Let
$$S(q,a,b) = (q;q)\_n \sum\_{i=0}^n\frac{(a;q)\_i}{(q;q)\_i}\frac{(b;q)\_{n-i}}{(q;q)\_{n-i}}a^{n-i} - (ab;q)\_n.$$
Your identity is equivalent to $S$ vanishing identically. Since $S$ is polynomial in its variables, by Lemma 1 it suffices to show that
$$(\*) S(q,q^A, q^B) = 0$$
for all $q,A,B$. Since $(q^A;q)\_i =\frac{(q;q)\_{i+A-1}}{(q;q)\_{A-1}}$, $(\*)$ is equivalent to
$$\sum\_{i=0}^{n} \frac{(q;q)\_{i+A-1}}{(q;q)\_{A-1} (q;q)\_i} \frac{(q;q)\_{n-i+B-1}}{(q;q)\_{B-1} (q;q)\_{n-i}} q^{A(n-i)} = \frac{(q;q)\_{n+A+B-1}}{(q;q)\_{A+B-1} (q;q)\_{n}},$$
or, in $q$-nomial notation,
$$\sum\_{i} \binom{i+A-1}{i}\_q \binom{n-i+B-1}{n-i}\_q q^{A(n-i)} = \binom{n+A+B-1}{n}\_q,$$
which is true by Lemma 2.




---


Lemma 1 may be proved, for instance, using the Combinatorial Nullstellensatz. Lemma 2 is proved just like the [q-Vandermonde identity](https://en.wikipedia.org/wiki/Q-Vandermonde_identity), only that the proof uses the identity $\prod\_{k=0}^{n-1} (1-q^k t)^{-1} = \sum\_{k} \binom{n+k-1}{k}\_q t^k$ instead of $\prod\_{k=0}^{n-1} (1+q^k t) = \sum\_{k} \binom{n}{k}\_q q^{\binom{k}{2}} t^k$.


