Is there a mathematical function $f$ that takes $n!$ as an input, and returns $n$?  For example,   $f(720) = 6$
I don't want to do this through trial and error, and the best way I have found so far was to start dividing from 1.     $n! = \text {a really big number}$   Ex. $n! = 9999999$   Is there a way to approximate n or solve for n through a formula of some sort?  Update (Here is my attempt):  Stirling's Approximation: $$n! \approx \sqrt{2 \pi n} \left( \dfrac{n}{e} \right ) ^ n$$  So taking the log:     $(2\cdot \pi\cdot n)^{1/2} \cdot n^n \cdot e^{-n}$       $(2\cdot \pi)^{1/2} \cdot n^{1/2} \cdot n^n \cdot e^{-n}$       $.5\log (2\pi) + .5\log n + n\log n \cdot -n \log e$       $.5\log (2\pi) + \log n(.5+n) - n$   Now to solve for n:     $.5\log (2\pi) + \log n(.5+n) - n = r$       $\log n(.5+n) - n = r - .5 \log (2\pi)$   Now I am a little caught up here.
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