q-gamma function

Remarks

The products converge because $\sum q^k < \infty$, and $(q^x; q)_\infty \ne 0$ because $q^{x+k} \ne 1$ for all $k \ge 0$.

Example: $\Gamma_q(1) = 1$, since the two products are equal. Also $\Gamma_q(2) = (1-q)^{-1} \frac{(q;q)_\infty}{(q^2;q)_\infty} = \frac{1 - q}{1 - q} = 1$ and $\Gamma_q(3) = \frac{(1-q)(1-q^2)}{(1-q)^2} = 1 + q$. In general $\Gamma_q(x + 1) = [x]_q\, \Gamma_q(x)$ with $[x]_q = \frac{1 - q^x}{1 - q}$, so $\Gamma_q(n + 1) = [1]_q [2]_q \cdots [n]_q$.

Non-example of a valid argument: for $x = -1$ the factor $1 - q^{x+1}$ of $(q^x; q)_\infty$ is $0$, so $\Gamma_q$ has a pole there.

For each fixed $x$, $\Gamma_q(x) \to \Gamma(x)$ as $q \to 1^-$, where $\Gamma$ is the gamma function. A related definition exists for $q > 1$.