Two problems on number theory I need some ideas (preferable some tricks) for solving these two problems:

Find the largest number $n$ such that $(2004!)!$ is divisible by
  $((n!)!)!$
For which integer $n$ is $2^8 + 2^{11} + 2^n$ a perfect square?

For the second one the suggested solution is like this  : $ 2^8 + 2^{11} + 2^n = ((2^4)^2 + 2\times2^4\times2^6 + (2^ \frac{n}{2})^2 ) \Rightarrow n=12$
But I can't understand the approach,any ideas?

If $n=7$ then $n!=5040\gt2004$, so....
If $n=6$ then $n!=720\lt2004$, so....
