If $b\in\mathbb{Z}_{\geqslant2}$, is the number given by the following sum:
$M=\displaystyle\sum_{n\in\mathbb{Z}_{\geqslant0}}b^{-10^{n}}$
a Liouville number?
It has lots of zeros, for sure, and by examining the continued fraction for $M$ when $b=10$ has lots of large terms consisting of all nines. For example ($9_a$ denotes a number consisting of $a$ nines):
$\displaystyle\sum^{\infty}_{n=0}10^{-10^{n}}=\left[0~|~9,1,9_8,10,9_{80},1,9,9_8,1,9,9_{800},1,8,1,9_8,9,1,9_{80},10,9_8,1,9,9_{8000},\ldots\right]$
This continued fraction exhibits the same "mirroring" effect as Liouville's Constant where the terms after the next incrementally largest term that consists of all nines, $a_n$, are the same as the non-zero terms before it, except that the preceding term $a_{n-1}$ is replaced by a $1$ followed by $a_{n-1}-1$.
The definition of a Liouville number is "a real number x with the property that, for every positive integer n, there exist integers p and q with q > 1 and such that $0 < \left| x-\dfrac{p}{q} \right| < \dfrac1{q^n} $ or $q^n < \dfrac1{\left| x-\dfrac{p}{q} \right|} $.
In your case,
$\left| x-\sum_{k=1}^{n-1}\dfrac1{b^{10^{k}}}\right| \approx \dfrac1{b^{10^{n}}} $.
Here, $q = b^{10^{n-1}} $ so you want $(b^{10^{n-1}})^n \lt b^{10^{n}} $ or $b^{n10^{n-1}} \lt b^{10^{n}} $ which is not true for $n > 10 $.
Therefore your number is not a Liouville number.
That is why using $n!$ instead of $10^n$ works - try it.