# How do you find the nth root of n!?

##### 1 Answer

Dec 31, 2016

#(n!)^(1/n) ~~ (2pin)^(1/(2n))(n/e)(1+1/(12n)+1/(288n^2)-139/(51840n^3))^(1/n)#

#### Explanation:

The precise value requires you to calculate

#{ (0! = 1), (n! = n*(n-1)!) :}#

then take the

However, note that a good approximation to

#n! ~~ sqrt(2pin)(n/e)^n(1+1/(12n)+1/(288n^2)-139/(51840n^3)+O(n^(-4)))#

Then:

#(n!)^(1/n) ~~ (2pin)^(1/(2n))(n/e)(1+1/(12n)+1/(288n^2)-139/(51840n^3))^(1/n)#

This will be good for about

See also https://socratic.org/s/aAV5WyiQ for an alternative formula for