= Solution
Let $F=\mathbb F_q$. The <general linear group over a finite field> consists of all invertible <linear maps> $F^n\to F^n$, and the <special linear group over a finite field> is its determinant-one <subgroup>. Counting the ordered <bases> gives
$$
|GL_n(q)|=\prod_{i=0}^{n-1}(q^n-q^i)
=q^{n(n-1)/2}\prod_{j=1}^{n}(q^j-1).
$$
The <determinant> map is onto $F^\times$, since $\operatorname{diag}(t,1,\ldots,1)$ has <determinant> $t$. Its kernel is $SL_n(q)$, so
$$
\boxed{|SL_n(q)|=\frac{|GL_n(q)|}{q-1}.}
$$
For $n>1$, commuting with all the <elementary transvection matrices> $I+tE_{ij}$ forces a <matrix> to be scalar. Therefore the <center of a group> of $SL_n(q)$ consists of $\lambda I$ with $\lambda^n=1$. The multiplicative group of a <finite field> is cyclic, so this <center of a group> has size $d=\gcd(n,q-1)$. The <projective special linear group over a finite field> consequently has order
$$
|PSL_n(q)|=\frac{q^{n(n-1)/2}\prod_{j=2}^n(q^j-1)}{\gcd(n,q-1)}.
$$
Here is a proof of simplicity which isolates all the required ingredients. First we prove the <Iwasawa simplicity lemma> in the form needed. Suppose a group $H$ acts faithfully and primitively, is a <perfect group>, and has an abelian <subgroup> $U$ normal in a <stabilizer subgroup> $H_\alpha$, whose conjugates generate $H$. For a nontrivial <normal subgroup> $N$, its orbits form a <block system>. Faithfulness prevents all these orbits from being singletons, so primitivity makes $N$ transitive. Thus $H=NH_\alpha$. Modulo $N$, each conjugate of $U$ has exactly the same image as $U$, because we may write its conjugating element as $nh$ with $n\in N$ and $h\in H_\alpha$. The quotient $H/N$ is generated by this one abelian image and hence is abelian. Since $H$ is a <perfect group>, $H/N=1$. Thus every nontrivial <normal subgroup> is $H$.
Apply this to the action on <projective points>, that is, one-dimensional <vector subspaces> of $F^n$. Any ordered pair of distinct lines can be sent to any other by a <linear map> obtained by extending representatives to <bases>. Multiplying one target <basis> <vector> by a suitable nonzero scalar adjusts the <determinant> to one without changing either target line. Thus $SL_n(q)$ is <two-transitive> on these points. A <linear map> fixing every line is scalar: it is diagonal on a fixed <basis>, and fixing the lines spanned by $e_i+e_j$ makes all diagonal entries equal. The projective quotient therefore acts faithfully, two-transitively, and primitively.
For the line $\langle e_1\rangle$, use
$$
U=\{I+e_1f:f\in (F^n)^*,\ f(e_1)=0\}.
$$
These <matrices> multiply by addition of $f$, so $U$ is abelian. Conjugating by a <matrix> preserving $\langle e_1\rangle$ preserves $U$, so it is normal in that <stabilizer subgroup>. Its conjugates include every <elementary transvection matrix> $I+tE_{ij}$.
These <elementary transvection matrices> generate $SL_n(q)$ by row elimination. For completeness, row addition is elementary, signed row interchange is elementary, and compensating diagonal scalings are elementary because, in a two-dimensional coordinate block,
$$
w(t)=
\begin{pmatrix}1&t\\0&1\end{pmatrix}
\begin{pmatrix}1&0\\-t^{-1}&1\end{pmatrix}
\begin{pmatrix}1&t\\0&1\end{pmatrix}
=\begin{pmatrix}0&t\\-t^{-1}&0\end{pmatrix},
\qquad
w(t)w(-1)=\operatorname{diag}(t,t^{-1}).
$$
Elimination using these operations reduces a determinant-one <matrix> to the identity.
It remains to show that these groups are <perfect groups>. In the following <matrix> identities use $[x,y]=xyx^{-1}y^{-1}$; the alternative <group commutator> convention generates the same <commutator subgroup>. If $n\ge3$, distinct indices give the <group commutator> identity
$$
[I+aE_{ij},I+bE_{jk}]=I+abE_{ik}.
$$
Every elementary generator is therefore a <group commutator>. For $n=2$ and $q>3$, choose $\lambda\in F^\times$ with $\lambda^2\ne1$, and let $D=\operatorname{diag}(\lambda,\lambda^{-1})$. Then
$$
[D,I+tE_{12}]=I+(\lambda^2-1)tE_{12}.
$$
As $t$ varies this supplies all upper <elementary transvection matrices>, and conjugation supplies the lower ones. Thus $SL_n(q)$, and its projective quotient, are perfect in the asserted cases. The preceding simplicity argument now applies.
The two exceptions are genuine: the projective actions identify $PSL_2(2)$ with $S_3$ and $PSL_2(3)$ with $A_4$, of orders 6 and 12 respectively. The first contains its <normal subgroup> of order 3; the second contains the normal Klein four group. We have proved
$$
\boxed{PSL_n(q)\text{ is simple for }n>1,\quad
(n,q)\notin\{(2,2),(2,3)\}.}
$$
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