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 $\map P n$ be the proposition:
 * $\ds \prod_{k \mathop = 1}^n \paren {1 + q^{k - 1} x} = \sum_{k \mathop \in \Z} \dbinom n k_q q^{k \paren {k - 1} / 2} x^k$

Basis for the Induction
$\map P 1$ is the case:

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

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

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

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

Induction Step
This is the induction step:

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

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

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

Some refer to it as the $q$-binomial theorem.