How many bases are there in $\mathbf Z_5^4$

Question

How many bases are there in $\mathbf Z_5^4$ ?

I've been asked to solve the above mentioned problem and had no clue how to do it. The only clear thing for me is that such base contains 4 vectors. Knowing that, I was able to count the total amount of sets which contain 4 vectors. Also I checked that we can obtain any element except 0 from $\mathbf Z_5$ by multiplying any other element by a certain scalar. For example $3*2=1$, $3*4=2$ , $3*1=3$, $3*3=4$, $3*5=0$, in $\mathbf Z_5$. A base is the largest, linearly independent subset of vectors from $\mathbf Z_5^4$,so I know that I have to combinatorially find the number of those sets, but I've no clue how.

Answer 1

Hint: How many choices are there for the first element in an ordered basis $(v_1,v_2,v_3,v_4)$ of $Z_5^4$?

Suppose now that you pick one of the possible choices $v_1$. How many choices are there for $v_2$?

Answer 2

You know how you create a basis in a vector space: choose one vector $v_1\ne 0$, then choose another vector $v_2\not\in span(v_1)$, then choose another vector $v_3\not\in span(v_1,v_2)$ etc.

In this case: in how many ways can $v_1$ can be chosen? Then, how many elements are in $span(v_1)$, and in how many ways can you choose $v_2$? Again, how many elements are in $span(v_1, v_2)$, and in how many ways can you choose $v_3$? Finally, do the same for $v_4$.