[proofplan] We count invertible matrices by counting their columns. An $n \times n$ matrix over $\mathbb{F}_q$ is invertible exactly when its ordered list of columns is an ordered basis of the [vector space](/page/Vector%20Space) $\mathbb{F}_q^n$. The first column may be any nonzero vector, and after $k$ linearly independent columns have been chosen, the next column may be any vector outside their $k$-dimensional span. Multiplying the number of choices at each stage gives the stated product. [/proofplan]

[step:Identify invertible matrices with ordered bases of $\mathbb{F}_q^n$] Let $V:=\mathbb{F}_q^n$, viewed as an $n$-dimensional vector space over $\mathbb{F}_q$, and let $M_n(\mathbb{F}_q)$ denote the set of all $n \times n$ matrices with entries in $\mathbb{F}_q$. For each matrix $A \in M_n(\mathbb{F}_q)$ and each $j \in \{1,\dots,n\}$, let $c_j(A) \in V$ denote the $j$th column of $A$. Define the column map \begin{align*} \Gamma: M_n(\mathbb{F}_q) \to V^n, \qquad A \mapsto (c_1(A),\dots,c_n(A)). \end{align*} This map is bijective, since an $n \times n$ matrix is uniquely determined by its ordered list of columns. We claim that $A \in GL_n(\mathbb{F}_q)$ if and only if $(c_1(A),\dots,c_n(A))$ is an ordered basis of $V$. Let \begin{align*} L_A: V \to V, \qquad x \mapsto Ax \end{align*} be the [linear map](/page/Linear%20Map) represented by $A$ in the standard basis $(e_1,\dots,e_n)$ of $V$. For every $j \in \{1,\dots,n\}$, we have $L_A(e_j)=c_j(A)$. If $A$ is invertible, then $L_A$ is a vector space isomorphism, so it sends the basis $(e_1,\dots,e_n)$ to the basis $(c_1(A),\dots,c_n(A))$. Conversely, if $(c_1(A),\dots,c_n(A))$ is a basis of $V$, then $L_A$ sends a basis to a basis, hence is an isomorphism, so $A$ is invertible. Therefore $|GL_n(\mathbb{F}_q)|$ equals the number of ordered bases of $V$. [/step]

[step:Count the possible choices for each successive column]We count ordered bases $(v_1,\dots,v_n)$ of $V$. Since $|\mathbb{F}_q|=q$, the vector space $V=\mathbb{F}_q^n$ has $q^n$ elements. For the first vector $v_1$, the only forbidden choice is $0$, so there are $q^n-1$ choices. Now suppose $k \in \{1,\dots,n-1\}$ and linearly independent vectors $v_1,\dots,v_k \in V$ have already been chosen. Define their span by \begin{align*} W_k:=\operatorname{span}_{\mathbb{F}_q}\{v_1,\dots,v_k\} \le V. \end{align*} Because $v_1,\dots,v_k$ are linearly independent, the coordinate map \begin{align*} \theta_k: \mathbb{F}_q^k \to W_k, \qquad (a_1,\dots,a_k) \mapsto \sum_{i=1}^{k} a_i v_i \end{align*} is bijective. Hence $|W_k|=|\mathbb{F}_q^k|=q^k$. The next vector $v_{k+1}$ must lie outside $W_k$, and every vector outside $W_k$ gives a linearly independent list $(v_1,\dots,v_k,v_{k+1})$. Thus there are $q^n-q^k$ choices for $v_{k+1}$.[/step]

[guided]We are building an ordered basis one vector at a time. The condition at stage $k+1$ is not that the next vector be nonzero by itself, but that it avoid all linear combinations of the previously chosen vectors. Start with $v_1$. Since $V=\mathbb{F}_q^n$ has $q^n$ elements and the zero vector cannot appear in a linearly independent list, $v_1$ has $q^n-1$ possible choices. Now assume that $k \in \{1,\dots,n-1\}$ and that $v_1,\dots,v_k$ have already been chosen linearly independently. Define the subspace \begin{align*} W_k:=\operatorname{span}_{\mathbb{F}_q}\{v_1,\dots,v_k\}. \end{align*} Because the chosen vectors are linearly independent, every element of $W_k$ has a unique expression as a linear combination of them. Equivalently, the map \begin{align*} \theta_k: \mathbb{F}_q^k \to W_k, \qquad (a_1,\dots,a_k) \mapsto \sum_{i=1}^{k} a_i v_i \end{align*} is bijective. Since $\mathbb{F}_q^k$ has $q^k$ elements, this gives $|W_k|=q^k$. The vector $v_{k+1}$ must not belong to $W_k$. If $v_{k+1}\in W_k$, then it is a linear combination of $v_1,\dots,v_k$, so the list is linearly dependent. If $v_{k+1}\notin W_k$, then no nontrivial relation among $v_1,\dots,v_k,v_{k+1}$ can exist: solving such a relation for $v_{k+1}$ would place $v_{k+1}$ in $W_k$, unless the coefficient of $v_{k+1}$ were zero, and then the independence of $v_1,\dots,v_k$ forces all remaining coefficients to be zero. Therefore precisely the vectors outside $W_k$ are allowed, and their number is \begin{align*} |V\setminus W_k|=|V|-|W_k|=q^n-q^k. \end{align*}[/guided]

[step:Multiply the independent choices to obtain the order] By the multiplication principle for successive finite choices, the number of ordered bases of $V$ is \begin{align*} (q^n-1)(q^n-q)(q^n-q^2)\cdots(q^n-q^{n-1}). \end{align*} Equivalently, \begin{align*} \prod_{k=0}^{n-1}(q^n-q^k). \end{align*} By the identification of $GL_n(\mathbb{F}_q)$ with the set of ordered bases of $V$ through columns, this is exactly $|GL_n(\mathbb{F}_q)|$. Hence \begin{align*} |GL_n(\mathbb{F}_q)|=\prod_{k=0}^{n-1}(q^n-q^k). \end{align*} This proves the theorem. [/step]