twer2

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)$

$n! = [n cdot (n-1) cdot (n-2) cdot ... cdot 1] spacespace ext{ with 'n' terms }$

$$f(n) in O(g(n)) implies g(n) in Omega(f(n))$$