<s>[INST] What number do you remove from $1!2!\cdots 99!100!$ to get a perfect square? 
The title says it all, there is a product, as shown above, one of the factorials must be removed and the product will make a perfect square. Which one?

For example, you could remove $54!$?
 [/INST] First of all, you can write every $(n!)$ as $(n-1)!\cdot n$.
However, for now, just do this for all the odd values of $n$ and you get
$$(2!)(2!)\cdot 3 \cdot(4!)(4!)\cdot 5\cdots (98!)(98!)\cdot 99\cdot 100!$$
Now call $(2!)(4!)\cdots (98!)=C$ to simplify and your expression is equal to
$$C^2\cdot 3\cdot 5\cdot 7\cdots 99\cdot (100!)$$
Write out $100!$ and that's equal to
$$C^2\cdot 3\cdot 5\cdot 7\cdots 99\cdot 2\cdot 3\cdot 4\cdots 100$$
Rearrange order to get
$$C\cdot 3\cdot 3\cdot 5\cdot  5\cdot 7 \cdot 7\cdots 99\cdot 2\cdot 4\cdot 6\cdots 100$$
Now call $3\cdot 5\cdot 7\cdots 99=D$ and that's equal to 
$$C^2D^2 2\cdot 4\cdot 6\cdots 100 = C^2D^2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 2\cdot 4\cdots 2\cdot 50$$
Group all the twos together and you see that's equal to $$C^2D^22^{50}\cdot 50!$$
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