Is there an inverse to Stirling's approximation? The factorial function cannot have an inverse, $0!$ and $1!$ having the same value. However, Stirling's approximation of the factorial $x! \sim x^xe^{-x}\sqrt{2\pi x}$ does not have this problem, and could provide a ballpark inverse to the factorial function. But can this actually be derived, and if so how? Here is my work:
$$
\begin{align}
y &= x^xe^{-x}\sqrt{2\pi x}\\
y^2 &= 2\pi x^{2x + 1}e^{-2x}\\
\frac{y^2}{2\pi} &= x^{2x + 1}e^{-2x}\\
\ln \frac{y^2}{2\pi} &= (2x + 1)\ln x - 2x\\
\ln \frac{y^2}{2\pi} &= 2x\ln x + \ln x - 2x\\
\ln \frac{y^2}{2\pi} &= 2x(\ln x - 1) + \ln x
\end{align}
$$
That is as far as I can go. I suspect the solution may require the Lambert W function.
Edit: I have just realized that after step 3 above, one can divide both sides by e to get
$$\left(\frac{x}{e}\right)^{2x + 1} = \frac{y^2}{2e\pi}$$
Can this be solved?

As $n$ increases to infinity we want to know roughly the size of the $x$ that satisfies the equation $x! = n$. By Stirling
$$
x^x e^{-x} \sqrt{2\pi x} \sim n
$$
Just focusing on $x^x$ a first approximation is $\log n / \log\log n$. Now writing $x = \log n / \log\log n + x_1$ and solving approximately for $x_1$, this time using $x^x e^{-x}$ we get
$$
x = \frac{\log n}{\log\log n} + \frac{\log n \cdot ( \log\log\log n + 1)}{(\log\log n)^2} + x_2
$$
with a yet smaller $x_2$ which can be also determined by plugging this into $x^x e^{-x}$. You'll notice eventually that the $\sqrt{2\pi x}$ is too small to contribute. You can continue in this way, and this will give you an asymptotic (non-convergent) serie (in powers of $\log\log n$). For more I recommend looking at De Brujin's book "Asymptotic methods in analysis". He specifically focuses on the case of $n!$ case in one of the Chapters (don't have the book with me to check).
