potato
because $n^n > n!$ implies $log(n!) = O(n cdot logn)$
because $n^n > n! implies log(n!) = O(n cdot logn)$
## Amortized Complexity
> $ ext{How } n! ext{ is } O(n^n)$
$n! = [n cdot (n-1) cdot (n-2) cdot ... cdot 1] spacespace ext{ with 'n' terms }$
$$f(n) in O(g(n)) implies g(n) in Omega(f(n))$$