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? 
 A: You are only showing the case $N/2$ is even, but missing when it is odd.
Alternative to your method, the shortcut is:
$$1!\cdot 2!\cdot 3!\cdot 4!\cdots (N-1)!\cdot N!=(1!)^2\cdot 2\cdot (3!)^2\cdot 4\cdots ((N-1)!)^2\cdot N=(1!3!\cdots(N-1)!)^2\cdot 2^{N/2}\cdot \left(\frac{N}{2}\right)!$$
When $\frac{N}{2}$ is even, removing $\left(\frac{N}{2}\right)!$ is sufficient, though there can be other solutions when $N$ is a multiple of $4$, because when $N=4M$, then $(N/2)!=(2M)!$ is even, then it is an iteration of the problem.
When $\frac{N}{2}$ is odd, there is no solution, because neither $2^{N/2}$ nor $(N/2)!$ nor both are square. For example, $6!$, $10!$ do not have solutions.
A: 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:
$$E^2 O^2 2 \cdot 4 \cdot 6 \cdots N = E^2 O^2 \cdot 2 \cdot (2 \cdot 2) \cdot (2 \cdot 3) \cdot (2 \cdot 4) \cdots (2 \cdot (N/2))$$
A: I don't understand why you feel the $N=8$ situation invalidates your solution. If you remove $(8/2)!$, you do get a product that is a square. That seems to me what you were asked to do: determine that removing $(N/2)!$ gives a product that is a square provided $N$ was divisible by $4$. 
The fact that removing $3!$ works is just an extra way to do it, and works because $4!/3!=4$ is itself a perfect square. 
Similarly if $N$ were $16$, then you can remove $8!$ by your solution. You will also be able to remove $9!$ though, because the ratio of those two is a perfect square.
A: If $N$ is a multiple of $4$, your proof correctly shows that you can remove the factor $\bigl({\large{\frac{N}{2}}}\bigr)!$. 

However, as Poon Levi points out, your argument doesn't prove that when $N$ is a multiple of $4$, the factor $\bigl({\large{\frac{N}{2}}}\bigr)!$ is the only factor which can be removed.

In fact, when $N$ is a multiple of $4$, you can also remove the factor


*

*$\bigl({\large{\frac{N}{2}}}-1\bigr)!$, whenever ${\large{\frac{N}{2}}}$  is a perfect square.$\\[4pt]$

*$\bigl({\large{\frac{N}{2}}}+1\bigr)!$, whenever ${\large{\frac{N}{2}}}+1$  is a perfect square.


Thus, 


*

*For $N=8$, since $4$ is a perfect square, you can remove either $3!$ or $4!$.$\\[4pt]$

*For $N=16$, since $9$ is a perfect square, you can remove either $8!$ or $9!$.


A possible conjecture . . .


*
If $N$ is a multiple of $4$, at most two removals are possible, and only one removal is possible, namely $\bigl({\large{\frac{N}{2}}}\bigl)!$, unless either ${\large{\frac{N}{2}}}$ or ${\large{\frac{N}{2}}}+1$ is a perfect square.


A further note . . .


*
Your argument doesn't prove that there are no solutions when $N$ is even but not a multiple of $4$.


For example, when $N=2$, you can remove $2!$, and when $N=14$, you can remove either $8!$ or $9!$.

