[proofplan] We study the determinant homomorphism from $GL_n(\mathbb{F}_q)$ to the multiplicative group $\mathbb{F}_q^\times$. Its kernel is $SL_n(\mathbb{F}_q)$, and its image is all of $\mathbb{F}_q^\times$ because every nonzero scalar $a$ occurs as the determinant of the diagonal matrix with entries $a,1,\dots,1$. The finite [group homomorphism](/page/Group%20Homomorphism) counting formula then gives $|GL_n(\mathbb{F}_q)|=|SL_n(\mathbb{F}_q)|(q-1)$, which is the desired identity. [/proofplan]

[step:Identify $SL_n(\mathbb{F}_q)$ as the kernel of the determinant homomorphism] Let $\det:GL_n(\mathbb{F}_q)\to \mathbb{F}_q^\times$ be the determinant map, where $\mathbb{F}_q^\times=\mathbb{F}_q\setminus\{0\}$ is the multiplicative group of nonzero elements of $\mathbb{F}_q$. This map is a group homomorphism because determinant is multiplicative: $\det(AB)=\det(A)\det(B)$ for all $A,B\in GL_n(\mathbb{F}_q)$. By the definition of the special linear group, or equivalently by [citetheorem:9740], $SL_n(\mathbb{F}_q)=\ker(\det:GL_n(\mathbb{F}_q)\to \mathbb{F}_q^\times)$. [/step]

[step:Show that every nonzero field element is a determinant]We prove that $\det:GL_n(\mathbb{F}_q)\to \mathbb{F}_q^\times$ is surjective. Let $a\in\mathbb{F}_q^\times$. Let $M_n(\mathbb{F}_q)$ denote the set of $n\times n$ matrices with entries in $\mathbb{F}_q$. Define $D_a\in M_n(\mathbb{F}_q)$ to be the diagonal matrix whose first diagonal entry is $a$ and whose remaining diagonal entries are $1$. Since $a\ne 0$, all diagonal entries of $D_a$ are nonzero, so $D_a$ is invertible. Hence $D_a\in GL_n(\mathbb{F}_q)$, and its determinant is \begin{align*} \det(D_a)=a\cdot 1\cdots 1=a. \end{align*} Thus every $a\in\mathbb{F}_q^\times$ lies in the image of the determinant map, so \begin{align*} \operatorname{im}(\det)=\mathbb{F}_q^\times. \end{align*} This also covers $n=1$: in that case $D_a$ is the $1\times 1$ matrix $(a)$, which lies in $GL_1(\mathbb{F}_q)$ and has determinant $a$.[/step]

[guided]We need to know the exact size of the image of the determinant map. Since the codomain is $\mathbb{F}_q^\times$, the strongest possible statement is surjectivity. Let $a\in\mathbb{F}_q^\times$ be arbitrary. Let $M_n(\mathbb{F}_q)$ denote the set of $n\times n$ matrices with entries in $\mathbb{F}_q$. Define $D_a\in M_n(\mathbb{F}_q)$ to be the diagonal matrix with first diagonal entry $a$ and all other diagonal entries equal to $1$. Because $a\ne 0$ and $1\ne 0$ in the field $\mathbb{F}_q$, every diagonal entry of $D_a$ is nonzero. A diagonal matrix over a field is invertible exactly when all of its diagonal entries are nonzero, so $D_a\in GL_n(\mathbb{F}_q)$. The determinant of a diagonal matrix is the product of its diagonal entries. Therefore $\det(D_a)=a\cdot 1\cdots 1=a$. Since $a\in\mathbb{F}_q^\times$ was arbitrary, every element of $\mathbb{F}_q^\times$ is hit by the determinant map. Hence $\operatorname{im}(\det)=\mathbb{F}_q^\times$. The same construction includes the edge case $n=1$. Then $D_a$ is simply the $1\times 1$ matrix $(a)$, it is invertible because $a\ne 0$, and its determinant is $a$.[/guided]

[step:Count the determinant fibers and solve for $|SL_n(\mathbb{F}_q)|$] For any group homomorphism $\varphi:G\to H$ between finite groups, each fiber over an element of $\operatorname{im}\varphi$ has cardinality $|\ker\varphi|$: if $\varphi(g)=h$, then $\varphi^{-1}(\{h\})=g\ker\varphi$. Thus $|G|=|\ker\varphi|\,|\operatorname{im}\varphi|$. Apply this to $\varphi=\det:GL_n(\mathbb{F}_q)\to\mathbb{F}_q^\times$. From the previous steps, $\ker\det=SL_n(\mathbb{F}_q)$ and $\operatorname{im}\det=\mathbb{F}_q^\times$. Because $\mathbb{F}_q$ has $q$ elements, its nonzero subset $\mathbb{F}_q^\times$ has $q-1$ elements. Therefore $|GL_n(\mathbb{F}_q)|=|SL_n(\mathbb{F}_q)|\,|\mathbb{F}_q^\times|=|SL_n(\mathbb{F}_q)|(q-1)$. Dividing by $q-1$ gives $|SL_n(\mathbb{F}_q)|=\frac{|GL_n(\mathbb{F}_q)|}{q-1}$. [/step]