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n^k = (e^k)^log n and the factorial grows faster than an exponential (product of growing factors vs. product of constant factors).
1
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solved Prove that (log n)! = O(n^k)
[ad_1]
n^k = (e^k)^log n and the factorial grows faster than an exponential (product of growing factors vs. product of constant factors).
1
[ad_2]
solved Prove that (log n)! = O(n^k)