Interesting property about finite products of $111..1$'s

Question

Denote by $[n]!$ the following finite product $1\cdot 11\cdot 111\cdot\dots \cdot \underbrace{111\dots1}_{n}$. Prove that $[n]![m]!$ divides $[n+m]!$

My efforts: I wrote the $[n]!$ in the following form: $$[n]!=1\cdot 11\cdot 111\cdot\dots \cdot \underbrace{111\dots1}_{n}=\dfrac{9\cdot 99\cdot 999\cdot\dots \cdot \underbrace{999\dots9}_{n}}{9^n}=\dfrac{(10-1)(10^2-1)\dots(10^n-1)}{9^n}$$

Let's consider $\dfrac{[n+m]!}{[n]![m]!}=\dfrac{\prod \limits_{i=1}^{n+m}(10^i-1)}{\prod \limits_{i=1}^{n}(10^i-1)\prod \limits_{i=1}^{m}(10^i-1)}=\dfrac{(10^{n+1}-1)(10^{n+2}-1)\dots (10^{n+m}-1)}{(10^1-1)(10^2-1)\dots (10^m-1)}$;

I was trying to use the following lemma: $10^a-1\mid 10^b-1$ if and only if $a\mid b$.

The only thing I need to prove is the following fact: If we have set $\{1,2,\dots, m\}$ then for any $n\in \mathbb{N}$ the translation set $\{n+1,n+2,\dots, n+m\}$ has the following interesting property: $\sigma(1)\mid n+1, \sigma(2)\mid n+2, \dots, \sigma(m)\mid n+m$ for some permutation $\sigma$ of the set $\{1,2,\dots, m\}$.

I guess that this fact is correct but I am not able to prove that.

Would be very grateful if somebody can show how to prove that.

Answer 1

A repunit is a series of 1's, So $R_8$ is 11111111.

The repunits are a product of 'algebraic roots' $A_n$, which divide the repunit $R_m$ if $n \mid m$. The Repunit 12 $R_{12} = A_2A_3A_4A_6A_{12}$, since the numbers 2,3,4,6,12 each divide 12.

What we are seeking is that in $[m]![n]!\mid [m+n]!$, that the powers of $A_x$ that divides n consecutive repunits, is not less than those that divide the first n, this is evaluated for the first n! values.

Since for any range of n places, there are at least n DIV q multiples of q, and perhaps one more, for every q from 2 to n. Therefore the statement is true.

Consider m=4, n=3

      Found                   Required
  1  2  3  4 | 5  6  7     1  2  3  4 | 1  2  3
     2     2 |    2           2     2 |    2       A_2
        3    |    3              3    |       3    A_3
           4 |                      4 |            A_4

So here we see that [7]! is divisible by [3]![4]!, and this is true for any case.

Answer 2

You may obtain a proof of your statement

$[n]![m]!$ divides $[n+m]!$

from the fact that the Gaussian binomial coefficients are polynomials with integer coefficients.

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Addendum A proof of the latter statement.

Consider the rational functions in the variable $q$ given by $$ \left[ \frac{n}{k}\right]_{q} = \frac{\prod_{i=n-k+1}^{n} (q^{i} - 1)}{\prod_{i=1}^{k} (q^{i} - 1)}. $$ One proves rather directly the following identities $$ \left[ \frac{n}{0}\right]_{q} = \left[ \frac{n}{n}\right]_{q} =1, \qquad \left[ \frac{n}{k}\right]_{q} = \left[ \frac{n-1}{k-1}\right]_{q} + q^{k} \left[ \frac{n-1}{k}\right]_{q} \quad\text{for $0 < k < n$}. $$ This implies via induction on $n$ that $$ \left[ \frac{n}{k}\right]_{q} $$ is a polynomial in $q$ of degree $k (n - k)$.