# How can I solve this Big $O$ exercise? [duplicate]

How can I prove that $n \log_2(n) ∈ O(log(n!))$ is true?

We start by supposing that $f(n)< c g(n)$ is true,

which means that $n \log_2(n) > c \log(n!)$ for all $n>n_0$ and $c>0$.

## marked as duplicate by Did, user91500, Claude Leibovici, Watson, Kamil JaroszMay 31 '16 at 13:32

• Not that it matters, but in $\log(n!)$, what is the base? And you will end up wanting to show that $n\log_2(n)$ is less than $c\log(n!)$ for sufficiently large $n$. – André Nicolas Apr 13 '16 at 18:49
• There are a lot of related questions, but you could first start with the usual trick: $\log(n!) = \log \prod_{j=1}^n j = \sum_{j=1}^n \log j > \sum_{j=n/2}^n \log j > \sum_{j=n/2}^n \log \frac{n}{2} > \dots$... – Clement C. Apr 13 '16 at 18:54