How many of the numbers in $A=\{1!,2!,...,2015!\}$ are square numbers? Problem
How many of the numbers in $A=\{1!,2!,...,2015!\}$ are square numbers?
My thoughts
I have no idea where to begin. I see no immediate connection between a factorial and a possible square. Much less for such ridiculously high numbers as $2015!$.
Thus, the only one I can immediately see is $1! = 1^2$, which is trivial to say the least.
 A: Only $1!$. For $n>1$, let $p$ be the greatest prime with $p\le n$. Between $p$ and $2p$ there is another prime, so $2p>n$. Therefore, $p$ occurs only once in the factorization of $n!$ and hence, $n!$ is not a square.
A: Hint: (for example) $13!, 14!, \dots , 25!$ are all nonsquare numbers because all of them are divisible by $13$ only once. (because $13$ is a prime)
Similarly, $17!, 18!, \dots, 33!$ are nonsquare numbers.
Go on like this.
A: Of all $n!$ only $0!$ and $1!$ are perfect squares.
http://mathforum.org/library/drmath/view/55881.html

To prove that a factorial bigger than 1 can't be a perfect square, 
  first think about breaking down the factorial into prime factors. For 
  example, 4! = 24, and 24 = 2*2*2*3. In order for any number to be a 
  perfect square, it must contain an even number of each prime factor.
  Now think about the largest prime less than or equal to n. Call that 
  number p. That number appears only once in the list of factors 
  1*2*3*4*5*...p...*n, so it can't appear an odd number of times 
  unless 2*p is also less than n.
Now there's a semi-famous theorem called Bertrand's Postulate, which 
  says that there is always a prime number between n and 2n. This is
  just  what you need to show that 2p isn't less than n - because if it
  were,  there would have to be a larger prime than p that is also less
  than n.

