Solving $n!=x$ for $n$. Trying to solve / approximate for $n$ in
$$n!=x\tag 1$$
where $x$ is given, I started with Stirling's approximation
$$n!\approx\sqrt{2\pi n}(n/e)^n\tag 2$$
Taking log and divide by $e$:
$$\frac1{2e}\ln(2\pi n)+\frac ne\ln(n/e)\approx \frac1e\ln x \tag3$$
Now in order to proceed, I neglected the left term with $\ln(2\pi n)$ and took the 0-th branch of Lambert $W$ assuming $x$ is large:
$$\ln n - 1 = \ln(n/e) = W(\tfrac ne\ln(n/e))\approx W(\tfrac1e\ln x) \tag4$$
Finally, adding $1$ and taking $\exp$ yields the approximation
$$n \approx \exp\left(1+W\Big(\frac1e\ln x\Big)\right) \tag5$$
That approximation is not too bad, but the results are too big by ca. 0.7 for bigger numbers; for example $x=100!$ yields $n\approx 100.7$.
Question: Is there a way to improve the result? For example, something that's smarter than just throwing away $\ln(2\pi n)$ in $(3)$.
I tried to find a better approximation of $\sum_k \ln k$ but all I found was solutions that effectively referred back to Stirling's $(2)$.

That linked question doen't answer my question. First, it doesn't explain how to improve my calculation; and my question is about improving my calculation.  Second, the term +0.5 is coming out of the blue and without explanation, except that it's "good". So no, that answer didn't help. In particular I don't seek for sites to copy-paste from, I am striving to getting better technique etc.

I checked that other answer again and i still does not explain where the 0.5 is coming from.  It just asserts that value from the start and claims it it "good" starting with it.
 A: 
something that's smarter than just throwing away $\ln(2πn)$ in $(3)$

Taking up the comments, the first idea is to divide out $\sqrt{2\pi}$ from the start so that $(3)$ becomes:
$$\frac1{2e}\ln(n)+\underbrace{\frac ne\ln(n/e)}_{\textstyle=:f(n)}\approx \frac1e\ln\frac x{\sqrt{2\pi}} \tag{3'}$$
Second idea is to set $f(n) = \dfrac ne\ln(n/e)$ and to find $\Delta$ such that
$$f(n+\Delta)\approx f(n) + \Delta f'(n) \approx f(n) + \frac 1{2e}\ln n\tag{*}$$
i.e. one adds some amout $\Delta$ to the argument of $f$ such that it accounts for the term $(\ln n)/(2e)$ in the LHS of $(3')$ that was previously neglected. Solving $(*)$ for $\Delta$ gives:
$$\begin{align}
\Delta \approx \frac{\ln n}{2e\cdot f'(n)} 
= \frac{\ln n}{2e\cdot\Big(\dfrac1e(1+\ln (n/e))\Big)} = \frac12\\
\end{align}$$
This turns $(3')$ into
$$\frac1{2e}\ln(n)+\frac ne\ln(n/e)
\approx \frac {n+0.5}e\ln\frac {n+0.5}e
\approx \frac1e\ln\frac x{\sqrt{2\pi}} \tag{3"}$$
The rest of the derivation is the same: Apply Lambert $W$:
$$\ln\frac {n+0.5}e \approx W\left(\frac1e\ln\frac x{\sqrt{2\pi}}\right) \tag{4'}$$
and finally solve for $n$:
$$\bbox[10px,border:1px solid]{n \approx \exp\left(1+W\left(\frac1e\ln\frac x{\sqrt{2\pi}}\right)\right) -\frac12 }\tag{5'}$$
The results are much more precise now, for example:
$$3! \mapsto 2.9905$$
$$100! \mapsto 99.99991$$
which is perfectly fine for my purposes.
