Order of finite unitary group 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 a paper of Wall (page 33), it is mentioned that the order of this group is $q^{(n^2-n)/2} \prod\limits_{i=1}^{n} (q^i - (-1)^i)$. 

How to prove this? 

Any help will be appreciated. 
 A: This question from 2010 was just listed as "active", apparently because someone (not myself) downvoted it today.  Anyway, rather than prolong the previous list of comments, I'll offer an explicit answer by pointing to an online resource that slightly predates Steinberg's 1968 AMS Memoir and is still a useful way to get into the details about Chevalley groups and their twisted analogues.   In 1967-68, Steinberg gave a course at Yale which was written up by listeners and published in mimeographed form by the math department there.   These typed notes are not as readable as typeset material and lack an index, but have been scanned in PDF format and placed on Steinberg's homepage at UCLA http://www.math.ucla.edu/~rst/
In particular, Section 11 deals with twisted groups, their Bruhat decompositions, and their orders.    One advantage of this broader approach via Chevalley groups is that it makes transparent the close analogy between the order formulas for finite unitary groups and for special linear groups of similar size.   In any case, as previous comments already indicate, the suggestion in the question that "This may be an easy exercise ..." is misguided.    But the order formula itself is standard and well studied from a variety of directions in the literature cited. 
