If N > 1, then N! is not a perfect square.

Indeed, let N > 1 and let p be the largest prime that is less than or equal to N. Then 2p cannot be less than or equal to N - otherwise, a prime q greater than p and less than 2p (such a q exists by Bertrand's theorem), would be a prime greater than p that is less than or equal to N. Thus p defined as above appears with an exponent of 1 in the prime decomposition of N! and therefore N! cannot be a perfect square if N > 1.