Recent content by Kartik Yadav

To prove that n log n is big oh of log(n!), I did: n log n <= C log(n!) n log n/ log(n!) <= C Let k = 1 n > k, so for n = 2 2 log 2 / log 2 <= C 2 <= C C is an element of [2, infinity) Taking C = 2 and k = 1 can we say, n log n <= 2 log(n!) and hence n log n is big oh of log(n!) ?