Motivation

Let ${\mathbb F}_q$ be the field with $q$ (a power of some prime number) elements. Then the order of $GL_n({\mathbb F}_q)$ is $$(q^n-1)(q^n-q)\cdots(q^n-q^{n-1}).$$ The fact that this order is divisible by $q^{n(n-1)/2}$, which is the order of the uni-triangular subgroup, plays a role in the proof of Sylow's Theorem, but I am interested in another topic. The order of $SL_n({\mathbb F}_q)$ is actually $$q^{n(n-1)/2}(q^n-1)(q^{n-1}-1)\cdots(q^2-1).$$ From the formula above, and the fact that $M_n({\mathbb F}_q)$ has cardinal $q^{n^2}$, we see that the probability that the polynomial $Det_n$ (the determinant) takes the value $1$ over $({\mathbb F}_q)^{n^2}$ is $$\frac1q(1-q^{-2})(1-q^{-3})\cdots(1-q^{-n}).$$ Amazingly, this probability is strictly less than $\frac1q$. The probability is the same for $Det_n=x$ if $x\ne0$, while it is larger than $\frac1q$ for $Det_n=0$.

Strangely enough, this probability has a non-trivial limit $$p(q)=\frac1q\prod_{m=2}^\infty\left(1-\frac1{q^m}\right)$$ as $n\rightarrow+\infty$ (large random matrices with entries in ${\mathbb F}_q$). We have $p(q)>0$ because $$\sum_{m=2}^\infty\frac1{q^m}<+\infty.$$

Question : Does this expression has a closed form ? In other words, is there a simplest formula than this infinite product.

Motivation

Let ${\mathbb F}_q$ be the field with $q$ (a power of some prime number) elements. Then the order of $GL_n({\mathbb F}_q)$ is $$(q^n-1)(q^n-q)\cdots(q^n-q^{n-1}).$$ The fact that this order is divisible by $q^{n(n-1)/2}$, which is the order of the uni-triangular subgroup, plays a role in the proof of Sylow's Theorem, but I am interested in another topic. The order of $SL_n({\mathbb F}_q)$ is actually $$q^{n(n-1)/2}(q^n-1)(q^{n-1}-1)\cdots(q^2-1).$$ From the formula above, and the fact that $M_n({\mathbb F}_q)$ has cardinal $q^{n^2}$, we see that the probability that the polynomial $Det_n$ (the determinant) takes the value $1$ over $({\mathbb F}_q)^{n^2}$ is $$\frac1q(1-q^{-2})(1-q^{-3})\cdots(1-q^{-n}).$$ Amazingly, this probability is strictly less than $\frac1q$. The probability is the same for $Det_n=x$ if $x\ne0$, while it is larger than $\frac1q$ for $Det_n=0$.

Strangely enough, this probability has a non-trivial limit $$p(q)=\frac1q\prod_{m=2}^\infty\left(1-\frac1{q^m}\right)$$ as $n\rightarrow+\infty$ (large random matrices with entries in ${\mathbb F}_q$). We have $p(q)>0$ because $$\sum_{m=2}^\infty\frac1{q^m}<+\infty.$$

Question : Does this expression has a closed form ? In other words, is there a simplest formula than this infinite product.

Edit. As mentionned by Pietro and Gerald (I realized that too, but was away from my internet connection), this limit probability can be expressed in terms of Dedekind's eta function: $$p(q)=\frac1{q^{1/24}(q-1)}\eta(\tau),\qquad q=e^{-2i\pi\tau}\hbox{ and }\Im\tau>0.$$ I apologize to those who use to denote $q$ the expression $e^{2i\pi\tau}$, the inverse of the present $q$. It is really unfortunate that it is used in field theory too. Now a related question:

What are the special values of Dedekind's function ? For instance, do we know a simplest expression for $q=2$ ?