What is the maximum value of $n$ if $4^n$ divides $1000!$ without a remainder? 
If $1000!$ is divided by $4^n$ with a remainder 0, what is the highest
  possible value of $n$?

I placed 2, 3, 4, etc value in $n$ but didn't found any possible $4^n$. Moreover I have seen that only $4^1$ can divide 1000! without remainder. Is there any way to prove this or there are some possible $n$? 

For a solution by hand: $1000!$ has 
$500$ multiples of $2^{1}$
$250$ multiples of $2^{2}$
$125$ multiples of $2^{3}$
$62$ multiples of $2^{4}$
$31$ multiples of $2^{5}$
$15$ multiples of $2^{6}$
$7$ multiples of $2^{7}$
$3$ multiples of $2^{8}$
$1$ multiples of $2^{9}$
So $2^{1+3+7+15+...+500}=2^{994}=4^{497}|1000!$
