Proving n log n is Big-Oh of log(n!)

[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!) ?

 

[BvU]

Hello Kartik,

You familiar with the Stirling formula ?

 

[Guneykan Ozgul]

Hi,
You can use Stirling's approximation as follows:
log(n)≅nlog(n)-n ∈O(nlog(n)) .