TITLE: $n!+1$ being a perfect square
QUESTION [53 upvotes]: One observes that 
\begin{equation*}
4!+1 =25=5^{2},~5!+1=121=11^{2} 
\end{equation*}
is a perfect square. Similarly for $n=7$ also we see that $n!+1$ is a perfect square. So one can ask the truth of this question:

Is $n!+1$ a perfect square for infinitely many $n$? If yes, then how to prove.

REPLY [45 votes]: This is Brocard's problem, and it is still open.
http://en.wikipedia.org/wiki/Brocard%27s_problem

REPLY [8 votes]: The sequence of factorials $n!+1$ which are also perfect squares is here in Sloane. It contains three terms, and notes that there are no more terms below $(10^9)!+1$, but as far as I  know there's no proof.

REPLY [4 votes]: My intuition would be that there are very few.  There are just not many squares and even fewer factorials.  OEIS A025494 lists the squares which are a sum of distinct factorials, which is less restrictive than what you ask and says the list is probably finite.  In particular, there are no more below 31!