Factorial Perfect Square Product

We can go through the prime factorization of each integer value of x. Since a prime will be multiplied by x! to yield a perfect square, we want exactly one of the prime factors to be of odd multiplicity. This is true for 2, 6, and 10. Ten is the largest with a prime factorization of: $2^{8} \times 3^{4} \times 5^{2} \times 7$ . We can confirm that prime factors of multiplicity one (odd) will be added too quickly to the factorization, given Bertrand's Postulate. Thus, our answer is $\boxed{10}$ , where 10! can be multiplied by 7 (prime) to give a perfect square.