I've been puzzling my puzzler about Sudoku solutions. The thing is, people already know how many solutions there are, but I don't. Rather than read about it, I'll just derive it here, or try until "bored" of it!
Counting in beautiful simplicity
I will start with the following board. It has got all the 1's and nothing else.
Hey, I see a cube! A skewered cube. No pun intended. If it was intended, I'd have simply said $ (x^{3})^{2} $!!!! So, I'm convinced that the above board is actually the only answer board. Other boards are just shuffled versions with more numbers included. Here's my thought process.
 I see each row in a family of 3 and each column in a family of 3. That's 6 groups in all.
 Next, I pivot rows and columns maintaining the families of 3.
 Lastly, I stare at it with my imagination until I see every puzzle's 1's placement accounted.
 I notice how each family of 3 rows has a small number of ways to pivot, and so does each family of 3 columns. That's $ 3! $ ways for each family.
 Because I have 6 independent families the product of pivots is $ (3!)^{6} $.
 Now that's a squared cube!
My first conclusion is that a skewered cube of 1s and its shuffles form a squared cube of possibilities.
Numbering 1s' locations
I can describe the cell location of each 1 with a vector.
 To prepare, I number the cells in a 3x3 block, proceeding top to bottom, left to right (column first) 1 4 7 2 5 8 3 6 9
 To create the vector, I proceed from block to block starting left to right, then top to bottom (row first ordering)
 For each block, I write down the number of the cell which contains a 1.
 With these steps, I form this vector from the picture above: [0, 1, 2, 3, 4, 5, 6, 7, 8]

For visualization, I rewrite this sequence in base 3 with line breaks (to match block location) and colors in column digit families:
00 01 02
10 11 12
20 21 22

Or colors in row digit families:
00 01 02
10 11 12
20 21 22  Swapping numbers in each of the 6 families above is equivalent to pivoting rows or columns in families of 3 [Counting in beautiful simplicity].
Numbering all digit locations
Beginning with the 1s, I can assign a prototypical numbering to each number by a simple rotation:
00 01 02 10 11 12 20 21 22 *** prototypical numbering of cells of 1
01 02 10 11 12 20 21 22 00 *** of 2
02 10 11 12 20 21 22 00 01 *** of 3
10 11 12 20 21 22 00 01 02 *** of 4
11 12 20 21 22 00 01 02 10 *** of 5
12 20 21 22 00 01 02 10 11 *** of 6
20 21 22 00 01 02 10 11 12 *** of 7
21 22 00 01 02 10 11 12 20 *** of 8
22 00 01 02 10 11 12 20 21 *** of 9
The above 9x9 matrix (base 3, modulo 9) represents a single solved Sudoku puzzle.
Let $ G_{b}^{n} $ be any such matrix of a valid Sudoku solution.
 $ 9 $ options to choose $ G_{1}^{1} $. This is the location of the 1 in the first block (topleft 3x3 square).
 $ \frac{9!}{9} $ options to dependently choose $ G_{1}^{n}, n>1 $. This is all remaining contents of block 1 (topleft 3x3 square)
 $ \frac{{3!}^{6}}{9} $ options to dependently choose $ G_{b}^{1}, b>1 $. This is the location of all remaining 1s.
00 01 02 10 11 12 20 21 22 *** of 1
01 02 10 11 12 20 21 22 00 *** of 2
02 10 11 12 20 21 22 00 01 *** of 3
10 11 12 20 21 22 00 01 02 *** of 4
11 12 20 21 22 00 01 02 10 *** of 5
12 20 21 22 00 01 02 10 11 *** of 6
20 21 22 00 01 02 10 11 12 *** of 7
21 22 00 01 02 10 11 12 20 *** of 8
22 00 01 02 10 11 12 20 21 *** of 9
 Dependently choose $ G_{b}^{2}, b>1 $. This is all remaining contents of block 2 (topmiddle 3x3 square)... How many options?
To be continued.