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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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This should be helpful in forming $\LaTeX$ expressions for you. – J. M. Sep 4 '11 at 9:52
Related... – J. M. Sep 4 '11 at 12:32
@Doug: I have formatted your question in $\LaTeX$. I suggest you look over the code so that you see how such equations are formatted. – mixedmath Sep 4 '11 at 15:04
Doug, should we add (English) words for you to select from, too?? – The Chaz 2.0 Sep 4 '11 at 15:18
@The Chaz Wow. You clearly do not understand why a website for math questions should have symbols to help aid the person asking a question. I do not even think you realize how it is very appropriate for a website like this to have the math symbols when posting a question. – Doug Sep 5 '11 at 7:59
up vote 10 down vote accepted

A good approximation for $n!$ is that of Stirling: $n!$ is approximately $n^ne^{-n}\sqrt{2\pi n}$. So if $n!=r$, where $r$ stands for "really large number," then, taking logs, you get $\left(n+\frac12\right)\log n-n+\frac12\log(2\pi)$ is approximately $\log r$. Now you can use Newton's method to solve $\left(n+\frac12\right)\log n-n+\frac12\log(2\pi)=\log r$ for $n$.

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Shouldn't it be n+(1/2)? in the first part of: (n−(1/2))logn−n+(1/2)log(2π). And what does taking the log of stirling's approximation mean? I honestly am a little bit unclear with logs when trying to isolate n. Could you show me step by step how to do that? – Doug Sep 4 '11 at 9:27
@Doug: you're right. Gerry, I hope you don't mind that I fixed your sign slip. – J. M. Sep 4 '11 at 9:35
I just added my attempt, hopefully someone can give me a hand. Also, should I start a new question for explanation of what is the log actually doing? – Doug Sep 4 '11 at 9:41
Thanks for fixing the sign. I made another mistake, writing $r$ where I needed $\log r$ at the end. I'll fix that one. Doug, all the log is doing is making the problem look a little nicer. You can't isolate $n$; what you can do, as I said, is use Newton's Method, which you can look up. Newton's Method uses Calculus. If you haven't done Calculus yet, you could start by seeing how close $n=\log r/\log\log r$ is to a solution, and then go a little higher or lower as necessary. – Gerry Myerson Sep 4 '11 at 12:31

$\exp(1+W(\log(n!)/e))-\frac{1}{2}\searrow n$ as $n\to\infty$where $W$ is the Lambert W function. For integer $n\ge2$, the floor is exact.

I just took the expansion out a bit more and the approximation I gave above overestimates $n$ by $\log\left(\sqrt{2\pi}\right)/\log(n)$.

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a better approximation involving $W$ as well : – Stéphane Laurent Apr 25 '15 at 10:11

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