What 1 number do you remove from $(1!)(2!)..(99!)(100!)$ to get a perfect sqaure

Group them as $$\begin{align*}(1!2!)(3!4!)\cdots (99!100!) &= (1!^2\times 2)(3!^2 \times 4)\cdots (99!^2\times 100) \\&= (1!3!\cdots 99!)^2 \times 2\cdot 4\cdot 6\cdots 100 \\ & = (1!3!\cdots99!)^2\times 2^{50} \times 50! \end{align*} $$

so you can remove $50!$ and it's a square.

Hint. Show that the number to remove is $50!$ (note that $50$ is half of $100$).

P.S. More generally, if $m$ is a positive integer then $$N:=\prod_{j=1}^{4m}(j!)=\prod_{k=1}^{2m}\left[(2k-1)!(2k)!\right]=\left(2^m\prod_{k=1}^{2m}(2k-1)!\right)^2\cdot (2m)!$$ which implies that $N/(2m)!$ is a perfect square.

What you need to do is keep track of unbalanced occurrences of prime factors. Since $(2n)!=(2n-1)!(2n)$, every pair of these consecutive factors $((2n-1)!, \ (2n)!)$ contributes the same as a single factor $2n$ to the unbalance of prime factors. Since we can pair up all $100$ factorials into $50$ such pairs, the unbalance is the same as that of the product $2\times 4\times\cdots\times 98\times 100$, which product equals $2^{50}50!$. Since $2^{50}$ is a perfect square, you can remove all unbalance by removing the factor $50!$ from the product.

Here is a tabular approach. Make up the table:

Note that in even columns the degree is even (e.g. $3^{98}$). From odd columns we will take away by one number and collect them to get $2\cdot4\cdots100=2^{50}\cdot50!$. Hence $50!$ must be removed to make the product a square.