Count invertible matrices by choosing an ordered basis one independent column at a time.Suggest a slogan
Let \(q\) be a prime power and \(n\geq1\). Then \[ \abs{\GL_n(\FF_q)} = \prod_{k=0}^{n-1}\qty{q^n - q^k} = (q^n-1)(q^n-q)\cdots(q^n - q^{n-1}). \]
A matrix in \(M_n(\FF_q)\) is invertible if and only if its columns form a basis of \(\FF_q^n\) [@DF04], so we count ordered bases. If the first \(k\) columns are linearly independent, the \((k+1)\)-st column can be any vector outside their span, a subspace with \(q^k\) elements, so there are \(q^n-q^k\) choices. Multiplying over \(k=0,\ldots,n-1\) gives the formula.