Gaussian Binomial Theorem

Theorem
Let $n \in \Z_{>0}$ be a (strictly) positive integer.

where $\dbinom n k_q$ denotes a Gaussian binomial coefficient.

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 1}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \prod_{k \mathop = 1}^n \left({1 + q^{k - 1} x}\right) = \sum_{k \mathop \in \Z} \dbinom n k_q q^{k \left({k - 1}\right) / 2} x^k$

Basis for the Induction
$P \left({1}\right)$ is the case:

Thus $P \left({1}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 1$, then it logically follows that $P \left({r + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\displaystyle \prod_{k \mathop = 1}^r \left({1 + q^{k - 1} x}\right) = \sum_{k \mathop \in \Z} \dbinom r k_q q^{k \left({k - 1}\right) / 2} x^k$

from which it is to be shown that:
 * $\displaystyle \prod_{k \mathop = 1}^{r + 1} \left({1 + q^{k - 1} x}\right) = \sum_{k \mathop \in \Z} \dbinom {r + 1} k_q q^{k \left({k - 1}\right) / 2} x^k$

Induction Step
This is the induction step:

So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_{\ge 1}: \displaystyle \prod_{k \mathop = 1}^n \left({1 + q^{k - 1} x}\right) = \sum_{k \mathop \in \Z} \dbinom n k_q q^{k \left({k - 1}\right) / 2} x^k$

Also known as
Some sources give this as the $q$-nomial theorem.