**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 $q=2$ ?