Comments on A070932.
Eric M. Rains, Jan 07 2013.

Theorem. A070932 (with 0 removed) is the multiplicative monoid generated by all numbers of the form q^n-q^{n-1} for n\ge 1 and q a prime power.  

Proof.
Call this monoid U, and note that the example |\F_q[t]/t^n|=q^n-q^{n-1}) shows that U is contained in A070932.

Lemma. Any number of the form |\GL_n(\F_q)|q^{ln} is in U.

Proof.  We have

|\GL_n(\F_q)|q^{ln}
=
(q^n-1)q^{ln}
|\GL_{n-1}(\F_q)|q^{n-1}.
QED

     Let us replace A070932 by the possibly larger monoid of numbers of the form |R^*||M| where R is a finite unital ring and M a finite left R-module; I claim that this is also equal to U.

     Let J=\rad R be the Jacobsen radical of R.  Since R is artinian, J is nilpotent (see Chapter 2 of Lam's "A First Course in Noncommutative Rings"), and thus

R^* = (1+J)\semidirect (R/J)^*.

Now, the subgroup 1+J has a filtration

1+J\supset 1+J^2\supset 1+J^3\supset\cdots

of normal subgroups, and since J is nilpotent, this filtration is finite.  The successive quotients of this filtration are just the additive groups J^k/J^{k-1}, each of which is in a natural way a module over R/J.

|R^*| = |(R/J)^*| |M_1|

where M = \bigoplus_{k\ge 1} J^k/J^{k+1} is a module over R/J.  Similarly,

|M| = |M_2|

where M_2 = \bigoplus_{k\ge 0} J^k M/J^{k+1}M is another R/J-module.

     In other words, we obtain the same set of numbers if we restrict to the case that J=0, i.e., that R is semisimple.  If we write R as a product of simple rings, that decomposition carries over to M, and thus the numbers with R simple generate the monoid.  In other words, the |R^*||M| monoid is generated by the numbers corresponding to R=\Mat_n(\F_q).  In that case, M is a finite sum of copies of the fundamental module, and thus |M|=q^{ln}; by the lemma, those numbers are in U.

QED