Where’s the error in my factorial proof? On one of our tests, the extra credit was to find which number you would take out from the set $\{1!,2!,3!,...(N-1)!,N!\}$ such that the product of the set is a perfect square, for even $N$ My answer was as follows:
Assume $N$ is even. First note that  $(n!)=(n-1)!\cdot n$.
Apply this to the odd numbers to get the product:
$$(2!)(2!)\cdot 3 \cdot(4!)(4!)\cdot 5\cdots ((N-2)!)((N-2)!)\cdot (N-1)\cdot N!$$
Let $ 2!4!6!...(N-2)!=E$. Then our equation is equal to: 
$$E^23\cdot 5\cdot 7\cdots (N-1)\cdot (N!)$$
Expand $N!$:
$$E^2\cdot 3\cdot 5\cdot 7\cdots (N-1)\cdot 2\cdot 3\cdot 4\cdots N$$
Group the odd terms together:
$$E^2\cdot 3\cdot 3\cdot 5\cdot  5\cdot 7 \cdot 7\cdots (N-1)\cdot 2\cdot 4\cdot 6\cdots N$$
Let $O=1\cdot3\cdot5...\cdot (N-1)$:
$$E^2O^2 2\cdot 4\cdot 6\cdots N = E^2O^2\cdot (2\cdot 2)\cdot (2\cdot 3)\cdot (2\cdot 4)\cdots (2\cdot (N/2))$$
Group together the $2$'s:
$$E^2O^22^{(N/2)}1\cdot2\cdot3...(N/2)=E^2O^22^{(N/2)}\cdot(N/2)!$$
So, if $N/2$ is even, it can be expressed as $2m$ for some $m$. So we have:
$$E^2O^22^{2m}(N/2)!=(EO2^m)^2(N/2)!$$
Therefore, if $N$ is even, the number missing is $(N/2)!$ if $N/2$ is even. For example, for $N=4$, $2!$ is missing, $N=6$ is impossible ($3$ is odd), and for $N=100$, $50!$ is missing.
However, this turned out not to be correct - indeed, for $N=8$ we have solutions of $3!$ and $4!$. So where did I go wrong in my proof, or what did I leave out? 

On this line:
$$E^2 O^2 2 \cdot 4 \cdot 6 \cdots N = E^2 O^2  \cdot (2 \cdot 2) \cdot (2 \cdot 3) \cdot (2 \cdot 4) \cdots (2 \cdot (N/2))$$
I believe you missed the first $2$. It should be:
$$E^2 O^2 2 \cdot 4 \cdot 6 \cdots N = E^2 O^2 \cdot 2 \cdot (2 \cdot 2) \cdot (2 \cdot 3) \cdot (2 \cdot 4) \cdots (2 \cdot (N/2))$$
