Let $S = 1! 2! \dotsm 100!$ Prove that there exists a unique positive integer $k$ such that $S/k!$ is a perfect square.
I've seen this question asked before but the answers were quite confusing. Anyone have a simpler solution to this problem? I believe the idea is to factor out perfect squares, but I'm not entirely sure how this works.
Here's an easy answer for the first part
$S = (1!) (1! \cdot 2) (3!) (3! \cdot 4) \cdots (99!) (99! * 100)$
$=(1!) (1! \cdot 2\cdot 1) (3!) (3! \cdot 2\cdot 2) \cdots (99!) (99! * 2\cdot 50)$
$=(1!)^2(3!)^2\cdots(99!)^2\cdot 2^{50}\cdot (50!)$
It is now easy to see that $\frac{S}{50!} = [(1!3!\cdots99!)(2^{25})]^2$