n^k = (e^k)^log n
and the factorial grows faster than an exponential (product of growing factors vs. product of constant factors).
1
solved Prove that (log n)! = O(n^k)
n^k = (e^k)^log n
and the factorial grows faster than an exponential (product of growing factors vs. product of constant factors).
1
solved Prove that (log n)! = O(n^k)