This may be an easy exercise but I am not getting it. Let $\mathbf F_q$ be a finite field with $q$ elements and $\mathbf F_{q^2}$ be its degree two extension. Define an automorphism $\sigma$ of $\mathbf F_{q^2}$ by $\sigma (x) = x^q$. For any matrix $A = (a_{ij}) \in M_n(\mathbf F_{q^2})$, let $A^{\star} = (a_{ji}^{\sigma})$ (i.e., $A^{\star} = A^{\sigma \mathrm{t}}$). Then finite unitary group $U_n(q)$ is a given by

$ U_n(q) = \left\lbrace A \in GL_n(\mathbf F_{q^2}) | A A^{*} = I_{n} \right\rbrace $

In paper of Wall (page 33), it is mentioned that order of this group is $q^{(n^2-n)/2} \prod\limits_{i=1}^{n} (q^i - (-1)^i)$. Question is how to prove this? Any help will be appreciated.

Geometry of Classical Groups over Finite Fields, second ed., Science Press, Beijing/New York, 2002. – Richard Stanley Aug 3 '10 at 14:56reference-request(for an old, standard result). The unified Lie-theoretic viewpoint is best, but there are nice older treatments such as Emil Artin's "The orders of the classical simple groups", Comm. Pure Appl. Math. 8 (1955). That was the year of Chevalley's famous Tohoku paper, to which Artin alludes. Calculating such orders of finite linear groups is not a trivial exercise, since it requires a strategy, but is by now fairly elementary. – Jim Humphreys Aug 3 '10 at 15:08