How many different Shidoku Boards are there? A Shidoku board is a 4 x 4 grid of numbers where each of the numbers 1-4   
appears exactly once in each row, column, and in each of the four 2 x 2 sub-
grids. How many different shidoku boards are there?"

For the first row there are $4!$ different possibilities. Second row there are $4$ different possibilities. Now for the third row... the bottom-right sub-board row depends on the bottom-left sub-board top row.
Ex:
1 2 3 4
3 4 2 1
4 1 x x
The first $x$ couldn't be 3 or 2 since that would mess up the column and it can't be 4 or 1 since those numbers are already used. 
This is the point where I am hitting my mental wall and am not progressing in the problem.   

Besides factoring out a factor of $4!$, we can also factor out an additional factor of 4.
First, label the upper-right-hand box as $a,b,c,d$.
a b * *
c d ? ?
& ? ? ?
& ? ? ?

There are of course $4!$ ways to choose the values of a,b,c,d.  Now the * signs must be c and d, and we can assume by switching the last two columns that they are in this order.  Similarly, the & signs are b and d, and we can switch the last two rows to get them in this order.  We've extracted a factor of $4$ by switching the rows and columns in this way, and we're left with
a b c d
c d ? ?
b ? ? ?
d ? ? ?

One can now simply check that
a b c d
c d ? ?
b a ? ?
d c ? ?

has two solutions and
a b c d
c d ? ?
b c ? ?
d a ? ?

has one, thus there are in total
$$
4! \cdot 4 \cdot (2 + 1) = 288
$$
solutions.
$$
* \quad * \quad *
$$
Note 1.
This problem is closely related to the the problem of counting Latin squares.
The reason this is relevant is that counting the number of Latin Squares of order $n$ is a very difficult in general, having only been calculated up to $n = 11$.
Intuitively, one expects the number of Sudoku boards (of any size) to be similarly hard to compute, and sure enough, the problem is quite difficult: see here, here, here, and here for some references.
Note 2.
Interestingly, the number of $4 \times 4$ Latin squares is 576,
which is exactly twice our above calculation of $288$.
In other words, if you try to generate a random "Shidoku" board by just ignoring the boxes and dealing only with rows and columns, there's a $50 \%$ chance you'll end up with a valid board anyway.
