solution:

please check if i do it write or wrong

n!=n*(n-1)*(n-2)...*3*2*1

let n=4 so we have

n!=n*(n-1)*(n-2)*(n-3)

= (n^2 - n)(n-2)(n-3)

=(n^3 -2n^2 - n^2 + 2n)(n-3)

=(n^3 -3n^2+2n)(n-3)

=n^4 - 3n - 3n^3 -3n^2 +2n^2 -6n

n!=n^4 -3n^3 - n^2 - 9n

so its big O is

O(n!)=O(n^4 -3n^3 - n^2 - 9n)

=O(n^4)

putting n=4 , we have

O(n!)=O(n^n)

which proves that n! is not O(2^n)