Asymptotic relation between $n^{1/3}$ and $(\log n)!$

Question

What is the asymptotic relation between $n^{1/3}$ and $(\log n)!$?

I compared the graphs and of course, $\mathcal{O}(n^{1/3})=(\log n)!$. However, for a more concrete proof, I compared the $\log$ of both functions although, I am not sure if this is a correct approach (and why/why not?). What I compared, as a result were $$\log n,~\log((\log n)!)\le n\log\log n$$ but, this is not really getting me anywhere.

I do not know how to approach such questions, help would be appreciated! For some similar problems I used the $\log$ comparison approach and ended up with correct solutions but I suspect it might be a faulty approach.

Edit. This questions possibly has a typo and it shouldbe $\log (n!)$ instead of $(\log n)!$. However, I am just as lost in that case.

@JoséCarlosSantos I was confused too but the question mentions $(\log n)!$. Is it possibly a typo? — reyna, Oct 17 '21 at 10:22
It could mean $\Gamma(\log n)$ which is roughly $\frac{(\log n)^{\frac12+\log n}}n$. — nejimban, Oct 17 '21 at 10:28
@nejimban Possible, but I think it is a simpler question. I'd like the answer in case it is $\log (n!)$. — reyna, Oct 17 '21 at 10:30
@zaira: $\log(n!)\sim n\log n$, which is of course larger than $n^{\frac13}$. — nejimban, Oct 17 '21 at 10:31
See, e.g., https://math.stackexchange.com/questions/140961/why-is-logn-on-log-n — Gary, Oct 17 '21 at 10:34

score 1 · Accepted Answer · answered Oct 17 '21 at 14:25

Assuming no typo in the question, we probably need to extend the factorial with $\log(n)!=\Gamma(\log(n)+1)$. Then, using Stirling's Approximation, we have $$ \log(n)!\sim\sqrt{2\pi\log(n)}\frac{\log(n)^{\log(n)}}{e^{\log(n)}}\tag1 $$ while $$ n^\alpha=e^{\alpha\log(n)}\tag2 $$ Dividing $(1)$ by $(2)$ gives $$ \begin{align} \frac{\log(n)!}{n^\alpha} &\sim\sqrt{2\pi\log(n)}\frac{\log(n)^{\log(n)}}{e^{\log(n)(1+\alpha)}}\tag{3a}\\ &=\sqrt{2\pi\log(n)}\left(\frac{\log(n)}{e^{1+\alpha}}\right)^{\log(n)}\tag{3b} \end{align} $$ Therefore, for any $\alpha$, $\log(n)!$ will eventually be larger than $n^\alpha$.

Asymptotic relation between $n^{1/3}$ and $(\log n)!$

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