A repunit is a number that contains only the digit 1 under base 10. Define the $k$-th repunit to be $r(k):=\sum_{i=1}^{k} 10^{i-1}$. Show that $$\left(\prod_{k=1}^m r(k)\right)\left(\prod_{k=1}^n r(k)\right) \mid \prod_{k=1}^{m+n}r(k),\quad \forall m,n\in\Bbb N_+$$ where $a\mid b$ means $a$ divides $b$.
Here's what I tried so far: I assumed $m\le n$ and tried investigating $RHS/LHS$ which is $$\frac{\prod_{k=n+1}^{n+m}r(k)}{\prod_{k=1}^mr(k)}$$
Then I incorrectly thought $\{r(k)\}$ are pairwise coprime (which is clearly false say for $r(2), r(4)$), and I proceeded to incorrectly conclude that to show $\prod_{k=1}^mr(k)$ divides $\prod_{k=n+1}^{n+m}r(k)$, it suffices to show $\prod_{k=n+1}^{n+m}r(k)$ is divisible by each $r(k),k=1,\cdots,m$, which is easy to prove per se because for each $k\in\{1,\cdots,m\}$ there is a $s\in\Bbb N_+$ such that $sk\in\{n+1,\cdots,n+m\}$, and since $r(k)\mid r(sk)$, we see that the claim holds. However, when I realised the precondition that $\{r(k)\}$ are pairwise coprime is wrong, the claim becomes insufficient to prove the desired result.
I'll assume that $r(k)=\sum_{i=0}^{k-1}10^i$ as otherwise the divisibility does not always hold, see my comment to the question.
You already note that $\prod_{k=1}^mr(k)$ divides $\prod_{k=1}^{m+n}r(k)$ and the quotient is $\prod_{k=m+1}^{m+n}r(k)$. So it suffices to show that $\prod_{k=1}^nr(k)$ divides $\prod_{k=m+1}^{m+n}r(k)$. Of course $$r(k)=\frac{10^{k}-1}{10-1}=[k]_{10},$$ which is known as the $q$-analog of $k$ at $q=10$. Now you want to show that $[n]_q!:=\prod_{k=1}^n[k]_q$ divides $\tfrac{[m+n]_q!}{[m]_q!}$, or equivalently, that the $q$-binomial coefficient $$\binom{m+n}{n}_q:=\frac{[m+n]_q!}{[m]_q![n]_q!},$$ is an integer. This is an exercise in elementary algebra; you can show this for any integer $q>1$ in general by induction, with the identity $$\binom{a}{b}_q=q^b\binom{a-1}{b}_q+\binom{a-1}{b-1}_q,$$ and the fact that $[1]_q=1$ and $[0]_q=0$ for all $q$.