For how many integers n between 1 and 50, inclusive, is (n^2 - 1)!/(n!

The numerator is
\((n^2 - 1)! = 1 * 2 * 3 * ... * n * (n+1) * (n+2) * ... * (2n) * ... * (3n) * ... * (n-1)*n * ... * (n^2 - 1)\)
The last number is 1 less than \(n^2 = n*n\)

e.g. n = 7

\((7^2 - 1)! = 1 * 2 * 3 * ... * 7 * 8 * 9 * ... 14 * ... 21 * ... 42 * ... * 48\)

The denominator is \((n!)^n = n! * n! * ... (n times)\)

Whenever n is prime, the numerator will have only (n-1) n's while the denominator will need n n's. Since n is prime, you cannot get n by multiplying any other two numbers. Hence, whenever n is prime, the given expression will not be an integer.

There are 15 prime numbers in the first 15 numbers so we are left with 35 numbers.

Check for squares of prime numbers too.

n = 4
\(\frac{(n^2-1)!}{(n!)^n} = \frac{(4^2-1)!}{(4!)^4} = \frac{1*2*3 .. 15}{(1*2*3*4)^4}\)

15! has 7+3+1 = 11 2s
4! has 2 + 1 = 3 2s
So 4 times 4! has 12 2s

Hence numerator has fewer 2s than denominator. This will not be an integer.

n = 9
\(\frac{(n^2-1)!}{(n!)^n} = \frac{(9^2-1)!}{(9!)^9} = \frac{1*2*3 .. 80}{(1*2*3*...*9)^9}\)

80! has 26+8+2 = 36 3s
9! has 3+1 = 4 3s
So 9 times 9! will have 36 3s
This should give us an integer.

Check for cube of a prime number

n = 8

\(\frac{(n^2-1)!}{(n!)^n} = \frac{(8^2-1)!}{(8!)^8} = \frac{1*2*3 .. 63}{(1*2*3*...*8)^8}\)

63! has 31+15+7+3+1 = 57 2s
8! has 4+2+1 = 7 2s
So 8 times 8! will have 56 2s
The expression will be an integer.

So in all, we have 50 - 15 - 1 = 34 numbers which will give us an integer.