The answer is **no**

If n is prime, then its obvious that n! is not a perfect square, for in that case n∣n! but n2∤n!. If n is not a prime, then let the r'th prime pr be the largest prime which is less than n. Now, n!=1⋅2⋅3⋅…⋅pr⋅(pr+1)⋅(pr+2)⋅…⋅(pr+n−pr). We can show that pr does not divide any of the numbers pr+1,pr+2,…,n. This is because any number greater than pr which is divisible by pr must be of the form mpr where m≥2. But by Bertrand's postulate, pr+1<2pr, where pr+1 is the (r+1)th prime. Thus if any of the numbers pr+1,pr+2,…,n is divisible by pr then one of these numbers must be the (r+1)th prime. But this contradicts the fact that pr is the largest prime which is less than n. So none of the numbers pr+1,pr+2,…,n is divisible by pr. So, we see that pr∣n! but p2r∤n!. Hence, n! is not a perfect square. This also proves that n! actually has even number of divisors.