q-binomial theorem

Remarks

Proof idea: let $f(z) = \sum_n c_n z^n$ be the left side, with $c_n = \frac{(a; q)_n}{(q; q)_n}$. Comparing the coefficients of $z^n$ on both sides, with $c_n (1 - q^n) = c_{n-1} (1 - a q^{n-1})$, gives $(1 - z) f(z) = (1 - az) f(qz)$. Iterating this functional equation gives $f(z) = \frac{(az; q)_N}{(z; q)_N} f(q^N z)$, and $f(q^N z) \to f(0) = 1$ as $N \to \infty$. Special cases: $a = 0$ gives $\sum_n \frac{z^n}{(q; q)_n} = \frac{1}{(z; q)_\infty}$, and replacing $a$ by $a/b$ and $z$ by $bz$ and letting $b \to 0$ gives $\sum_n \frac{(-1)^n q^{\binom{n}{2}} a^n z^n}{(q; q)_n} = (az; q)_\infty$. With $a = q^{-N}$ and $z$ replaced by $-q^N z$, one obtains the finite form $\prod_{k=0}^{N-1} (1 + q^k z) = \sum_{k=0}^{N} q^{\binom{k}{2}} \binom{N}{k}_q z^k$ with the Gaussian binomial coefficients. The hypothesis $\abs{z} < 1$ is needed for $a \notin \set{1, q^{-1}, q^{-2}, \dots}$, since the series then has radius of convergence $1$.

Going deeper

For a student

This is the simplest summation formula for basic hypergeometric series, and many others follow from it. Together with the case $a = 0$ it gives Euler's two expansions of $1/(z; q)_\infty$ and $(z; q)_\infty$, which are generating functions for partitions into parts of a given kind. Its bilateral extension is Ramanujan's 1psi1 summation.