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.


This is Brocard's problem, and it is still open.

http://en.wikipedia.org/wiki/Brocard%27s_problem
