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Moved from a non-homework section, so missing the template

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!) ?

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!) ?