Given $n^2$ different numbers from a field to form a $n$-degree matrix, prove that the determinant can take at most $$\frac{(n^2)!}{(n!)^2}$$ values.
The number of different matrices is $(n^2)!$. So it suffices to prove that for every matrix, there are $(n!)^2$ matrices which have the same determinant.
For a given matrix, we can arrange its columns arbitrarily, and after each column arrangement, we can perform a row arrangement to it.
So under such operations, this matrix can produce $(n!)^2$ different matrices.
If we want the later matrix to have the same determinant as the given one, we should confirm that the compound odevity of the column arrangement and the row arrangement to be even. Thus, only $ \text{even}\times\text{even} $ and $ \text{odd}\times\text{odd} $ can produce the same determinant.
Therefore, I can only find $\frac{(n!)^2}{2}$ matrices that have the same determinant for each given matrix.
So my question is where are the "other" $\frac{(n!)^2}{2}$ matrices?
The intended question is about how many different determinants there are, not about how big any one of them is. The part in the blockquote is missing some key words; it should be "the number of possible determinants".
Now, to answer it: the other thing we can do is take transposes. No amount of permuting rows and columns will get us from $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ to $\begin{pmatrix}a&c\\b&d\end{pmatrix}$, but they have the same determinant.