If you take the log of all your values, you can then treat it just like a linear scale. Using base 10 logs, if your data ranges from 10 to 100,000, taking logs makes it range from 1 to 5. Then scale that as you have before, presumably giving 10 lines to each unit to make 40 lines in your plot. Then plot you * where the log of the value belongs, so if you have a value of 2,000, the log of that is 3.3 and you would plot it on the 24th line.
I suspect there is no natural sigmoid function that will "preserve" in some sense your original function $f$.
In all generality, you want to "squeeze" the real line ($\mathbb{R}$) into an (open) interval of the form $(x_1,x_2)$. Let's try to squeeze both the $x$-axis and the $y$-axis so that we may fit one graph into a rectangle. Let this rectangle be, for simplicity, $(-a,a)\times(-b,b)$ where $a$ and $b$ are positive.
Let $D$ be the domain of $f$ and $R$ be its codomain (for the sake of simplicity, let's think of $R$ as $\mathbb{R}$)
To squeeze the $x$-axis we really need a function $\varphi:\ (-a,a)\longrightarrow D$, so that when we graph $f\circ\varphi$ we'll need to see what that thing does only on the interval $(-a,a)$.
To squeeze also the $y$-axis, we need another function $\psi:R\longrightarrow(-b,b)$. The end result is then
\begin{equation}
\tilde{f}:=\psi\circ f\circ\varphi
\end{equation}
Its graph will be contained in the rectangle $(-a,a)\times(-b,b)$.
The tricky part, as you pointed out, is finding the "right" sigmoid function(s). One can try various cocktails. If $D=R=\mathbb{R}$ for instance, one could take
\begin{equation}
\varphi(x):=k\tan\left(\frac{\pi}{2a}x\right)\qquad\psi(x):=\frac{2b}{\pi}\arctan(x/k)
\end{equation}
to do the job. Varying the parameter $k$ allows for trying to "match" the original function (for biggish $k$). It really depends on what characteristics of $f$ you want to study.
[EDIT: I personally like viewing functions "wrapped" on a sphere- but it's only an aesthetic approach. Here i posed a question about it.]
Best Answer
In a sense, because a factorial is "locally exponential".
What does this mean? Factorials satisfy the recurrence relation
$$(n + 1)! = (n + 1) \cdot n!$$
while exponentials satisfy
$$b^{n + 1} = b \cdot b^n$$
for a given base $b$. Now suppose we have an $n$ that's pretty large. Then $n + 1 \approx n$. Then we have that
$$(n + 1)! \approx n \cdot n!$$
which is similar to the base-$n$ exponential's recurrence, i.e.
$$n^{n+1} = n \cdot n^n$$
In particular, we can even go out some number $k$ of increments in the same way, provided that $k$ is not too big, to get:
$$(n + k)! \approx n^k \cdot n!$$
Hence, over a range of $k$ not so large that any of them are comparable to $n$, $k \mapsto (n + k)!$ is approximately an exponential function! And thus if you plot $y = (n + x)!$ with a large $n$ and not too large range of $x$, you will get an approximately exponential curve, which then becomes likewise approximately linear in the logarithmic plot.