Irrationality of the concatenation of the rightmost nonzero digits in $n!$ Surfing the internet I bumped into a very interesting problem, which I tried to solve, but got no results.
The problem is following: let $h_n$ be the most right non-zero digit of $n!$, for example, $10!=3628800,$ so $h_{10}=8$. The task is to prove that decimal fraction $0,h_1h_2\ldots h_n\ldots$ is irrational.

To show that $0,h_1h_2\ldots$ is irrational, it's enought to show that the sequence $h_n$ is not eventually periodic.
Assume on contrary that $h_n$ is eventually periodic.
Then $h_n$ is eventually periodic as a sequence modulo $5$.
Let $\mathbb Q^+$ denote the multiplicative group of positive rationals.
The set of prime numbers constitutes a base of $\mathbb Q^+$ regarded as $\mathbb Z$-module.
Consequently, there exists a unique multiplicative group homomorphism $\lambda:\mathbb Q^+\to\mathbb Q^+$ such that:
$$\lambda(p)=\begin{cases}p&p\neq 5\\\frac 12&p=5\end{cases}$$
($p$ prime number).
Then we have $h_n\equiv \lambda(n!)\pmod{10}$ for each $n>0$, hence
$$h_n\equiv\lambda(n!)\equiv \prod_{k=1}^n\lambda(k)\pmod{5}$$
Note that $\lambda(k)=k$ if $5\nmid k$.
Lemma. Let $u_n(n\in\mathbb N)$ be a sequence in a multiplicative group $G$.
If $\prod_k u_k$ is eventually periodic, then $u_n$ is eventually periodic.
For if
$$\prod_{k=0}^{n+T}u_k=\prod_{k=0}^{n}u_k$$ for each $n\geq N$, then
$$\prod_{k=n+1}^{n+T}u_k=1$$
for each $n\geq N$, hence
$$\prod_{k=n+1}^{n+T}u_k=1=\prod_{k=n+2}^{n+1+T}u_k$$
from which $u_{n+1}=u_{n+1+T}$ for each $n\geq N$.
Thus if $h_n$ is eventually periodic, then also $\lambda(k)$ is eventually periodic in the group of units of $\mathbb Z\diagup 5\mathbb Z$, that's $\lambda(n+T)\equiv\lambda(n)\pmod 5$ for each $n\geq N$.
Write $T=5^vT'$ with $5\nmid T'$; for $n\geq N$ and $n>v$ we have $\lambda(5^n)\equiv 3^n\pmod 5$ and
$$\lambda(5^n+T)=\lambda(5^n+5^vT')=3^v\lambda(5^{n-v}+T')\equiv 3^v(5^{n-v}+T')\equiv 3^vT'\pmod 5$$
thus $T'\equiv 3^{n-v}\pmod 5$ for each $n$ large enought - a contradiction which concludes the proof.
