Q:

How many (unordered) bases does $\Bbb F_q^n$ have as a vector space over $\Bbb F_q$? 

Following the recommendation here to get this question out of the unanswered queue, I've changed this from a proof-verification question into an answer-your-own. 
Here's the question again in case someone has their own proof (which may be better than my own below):

How many (unordered) bases does $\Bbb F_q^n$ have as a vector space over $\Bbb F_q$? 

A:

We shall first consider how many different ordered bases $\Bbb F_q^n$ has. 
Recall that $|GL_n(\Bbb F_q)|=(q^n-1)(q^n-q)\cdots(q^n-q^{n-1})$. Each element of $GL_n(\Bbb F_q)$ represents a linear map that carries the standard (ordered) basis $\{e_1, e_2, \ldots, e_n\}$ to another ordered basis.  We can establish a bijection between the ordered bases of $\Bbb F_q^n$ with the elements of $GL_n(\Bbb F_q)$ by the rule 
$$\{v_1, \ldots, v_n\}\mapsto\text{the linear extension of the map that carries the standard basis to } \{v_1, \ldots, v_n\}$$
Thus we know how many ordered bases $\Bbb F_q^n$ has.  Now we define an equivalence relation on the set of all ordered bases of $\Bbb F_q^n$. We call two ordered bases equivalent if they are permutations of one another.  The quotient set under this equivalence relation is clearly in bijection with the set of all unordered bases of $\Bbb F_q^n$. Overmore, each equivalence class has the same number of elements, namely $n!$ since there are $n!$ permutations of any ordered basis.  Since the set of unordered bases is in bijection with the quotient set we have that the total number of unordered bases of $\Bbb F_q^n$ is:
$$\frac{(q^n-1)(q^n-q)\cdots(q^n-q^{n-1})}{n!}$$
This approach can be mirrored to conclude that the number of linearly independent subsets of order $k$ is:
$$\frac{(q^n-1)(q^n-q)\cdots(q^n-q^{k-1})}{k!}$$

