Is it possible to determine the number divisors of n! especially for large n? I read this paper by P. Erdos, page 2. I didn't understand it. How do I determine the number divisors of $n!$ ? I'd like an example application, for example if I want to determine the number divisors of $150!$ then what can I do?  
Thank you for any help.
 A: If you want to apply some of his ideas to a specific $n$, then I would suggest starting with the formula at the beginning of section 2 that gives the expression for $n!$ as a product of $p^{w_p(n)}$ terms (the prime factorisation). The formula for $w_p(n)$ is simply expressing the fact that every $p-th$ number ranging from $1,...,n$ has a factor of $p$ and every $p-th$ one of these has an additional factor of $p$, etc. (or count multiples of $p,p^2,p^3,...$ and apply GPIE since every multiple of $p^2$ is also a multiple of $p$, and so on).   
Then you get an example such as:
$20! = 2^{10+5+2+1}\cdot3^{6+2}\cdot5^{4}\cdot7^{2}\cdot11\cdot13\cdot17\cdot19 = 2^{18}\cdot3^{8}\cdot5^{4}\cdot7^{2}\cdot11\cdot13\cdot17\cdot19$
Once the prime factorisation has been determined, then $d(n!)$ is a product of the $(1+w_p(n))$ terms, since each divisor is essentially an independent choice of each of the prime exponents in the factorisation. Erdös has just taken the $log$ of this relation to get a summation instead of a product.
Then, for example
$d(20!) = 19 \cdot 9 \cdot 5 \cdot 3 \cdot 2^4 = 41040$  
