n! can be written as n(n-1)(n-2).....1.
$n! = n(n-1)(n-2)...1.$
Highest order term in the product is $n^{n}$.
In time complexity , the lower order terms do not contribute.
Thus , $n! = \Omega(n^{n}) => log(n!) = \Omega(log(n^{n})) = \Omega(nlog(n))$
More formally , $log(n!) = \Theta(nlog(n))$