Is it a square

$1!\times 2!\times ...\times 99!\times 100! \\=1^{100}\times 2^{99}\times ...\times 99^{2} \times 100^{1} \\=x^{2}\times 2\times 4\times ...\times 100 \\=x^{2}\times 2\times 50!$

where x is a certain positive integer.

$2\times 50!$ is obviously not a square, so $1!\times 2!\times ...\times 99!\times 100!$ is not a square.

Hint: What is the power of 47 that appears in the number?

For any natural number n > 1, n! cannot be a square. It is evident that 2! and 3! aren't squares. Bertrand's postulate says that there always is a prime p such that k < p < 2k. Thus, for every number n > 3, there is a prime p that appears exactly once in n! (which can easily be seen when one takes 2k to be the smallest even number greater than or equal to n).