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? 

It suffices to show that $\frac{n}{2}\log\frac{n}{2}\geq cn\log n$ for some $c>0$. This is equivalent to $\log\frac{n}{2}\geq c_1\log n$ for some $c_1>0$. We can use $c_1=\frac{1}{2}$.
Clearly, the condition $\log\frac{n}{2}\geq\frac{1}{2}\log n$ is same as $\log n\geq \frac{1}{2}\log n+\log 2$ and thus equivalent to $\frac{1}{2}\log n\geq \log 2$. This is same as $\log_2 n=\frac{\log n}{\log 2}\geq 2$. This is true for $n\geq 4$.
