
> 
> Prove that for all positive integers $ n$ different from
>  $ 3$ and $ 5$, $ n!$ is divisible by the number of its positive
>  divisors. 
> 
> 
> 


I tried some things,such as the number of divisors of $ n!$ is just $ \prod\_p a\_p$ where $ a\_p=1+\sum\_{j\ge 1}\left\lfloor\frac n{p^j}\right\rfloor\le\min\left(1+\frac n{p-1},n\right)$.



Prove $ n!$ is divisible by the number of its positive divisors
The sequence, number of divisors of $n$-factorial, is tabulated at the [OEIS](https://oeis.org/A027423). It says there that it divides $n!$ for all $n\ge6$, giving a reference to Florian Luca and Paul Thomas Young, On the number of divisors of $n!$ and of the Fibonacci numbers, Glasnik Matematicki, Vol. 47, No. 2 (2012), 285-293, MR3006627. DOI: 10.3336/gm.47.2.05. It gives a link to the article, but my computer couldn't find the site. 


This [link](https://web.math.pmf.unizg.hr/glasnik/47.2/47(2)-05.pdf) might work. 


