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

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 n! using the recursive definition:

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

then take the $n$th root.

However, note that a good approximation to n! is given by the Stirling series:

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 $10$ significant digits.

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