Is there a way to solve for an unknown in a factorial? 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.

I believe you can solve Sterling’s formula for $n$ by way of the Lambert $W$ function, but that just leads to another expression, which can not be directly evaluated with elementary operations. (Although, I wish scientific calculators included the Lambert $W$, at least the principal branch).
I’ve derived a formula which is moderately accurate. If you are rounding to integer values of $n$, it will provide exact results through $170!$, and perhaps further. The spreadsheet app on my iPad will not compute factorials beyond $170!$.
If you are seeking rational solutions for $n$, in other words, trying to calculate values of a sort of inverse Gamma function, it will provide reasonably close approximations for $n$ in $[2, 170]$, and as I remarked above, possibly greater.
My Clumsy Little Equation:

I also have derived a variant which is much more accurate for $n$ in $[2, 40\text{ish}]$, however it starts to diverge beyond that.
I initially derived these approximations for my daughter, who, while working on Taylor/Maclaurin series, wanted a quick way to check for potential factorial simplifications in the denominator of terms.
I couldn’t find any solutions, except the obvious path through Stirling’s formula and the Lambert $W$ function. If more competent mathematicians can find a tweak which can improve accuracy, please share it.
I apologize in advance, as I am new here (as a contributor, that is), and I am not yet allowed to directly embed an image. Hopefully, the linked one works.
