Proving n(log(n)) is O(log(n!)) I want to prove $n(\log(n)) \in O(\log(n!))$. I don't really understand how to prove this statement. From the definition, we would have that:
$\exists c > 0, \exists N$, so that $\forall n \geq N, c\log(n!) \geq n\log(n)$
From expanding the terms, I found that:
$\log(n!) = \log(1) + ... + \log(\frac{n}{2}) + ... + \log(n-1) + \log(n) > \log(\frac{n}{2}) + ... + \log(n)$
$\log(n!) > (\displaystyle\frac{n}{2})\log(\displaystyle\frac{n}{2})$
However, I don't know how to proceed from this point. I found in another site that they concluded that $\log(n!) > (\displaystyle\frac{n}{2})\log(\displaystyle\frac{n}{2}) \in O(\log(n!))$ and therefore $log(n!) \in O(\log(n!))$. However, I don't see why this would be true as we had not found such a constant, as we would have then that $\log(\displaystyle\frac{n}{2}) = \log(n) -\log(2) \neq \log(n)$.
What would be the best way to prove this big O statement? 

A sketch of solution:
Firstly we have $\frac{n}{2} > \sqrt{n} $ when $n > 4$, thus $$n! \geq (\sqrt{n})^{\dfrac{n}{2}} = n^{n/4}$$
Then we have $\dfrac{n\log n}{\log n!} \leq \dfrac{n\log n}{\log n^{n/4}} = \dfrac{n\log n}{\frac{n}{4}\log n} = 4$
