25. Given a positive integer $n \in[1,50]$. Then the number of $n$ that satisfy $\frac{\left(n^{2}-1\right)!}{(n!)^{n}} \in \mathbf{Z}$ is ( ).
(A) 31
(B) 32
(C) 33
(D) 34
(E) 35
25. D.

Notice that, $\frac{\left(n^{2}\right)!}{(n!)^{n+1}} \in \mathbf{Z}$,
$\frac{\left(n^{2}-1\right)!}{(n!)^{n}}=\frac{\left(n^{2}\right)!}{(n!)^{n+1}} \cdot \frac{n!}{n^{2}}$.
To make $\frac{\left(n^{2}-1\right)!}{(n!)^{n}} \in \mathbf{Z}$, it is sufficient that $\frac{n!}{n^{2}} \in \mathbf{Z}$, i.e., $\frac{(n-1)!}{n} \in \mathbf{Z}$.
If $\frac{(n-1)!}{n} \notin \mathbf{Z}$, then $n=4$ or $n$ is a prime number.
Also, there are 15 prime numbers in the range $n \in \mathbf{Z}, n \in[1,50]$, so the number of positive integers $n$ for which $\frac{(n-1)!}{n} \in \mathbf{Z}$ is $50-15-1=34$.