how to prove that log(n!) >= c n log(n) for some c >0?
$n! = n \cdot (n-1) \cdot ... \cdot 1 \ge n \cdot (n-1) \cdot ... \cdot (n/2) \ge (n/2)^{(n/2)}$

so

$\log(n!)≥c\cdot n\cdot\log n$ for $c \ge 1/2$