Prove that $\log(n!)\leq(\log(n))!$

Question

Prove that $\log(n!)\leq(\log(n))!$

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My attempt:

I read somewhere that $n\leq\log(n!)\leq(\log(n))!$. But when I used calculator $\log(n!)$ can not be less than or equal to $(\log(n))!$.

Can you explain it in a formal way, please?

Answer 1

Let $x=\log(n)$

$\log(n!)<\log(n^n)=n\log(n)=xe^x$
$(\log(n))!=x!$

and the results follow by Factorial grows faster than Exponential.

Answer 2

As I suggested, using Strirling approximation is a very loose way, one immediately gets $$\log(n!)\approx n\log n$$ and $$\log(n)! \approx (\log(n))^{\log(n)}.$$ Cleraly the second dominates the first term, giving the claim.