We can see that in the decimal system each of $12345679\times k$ $(k\in\mathbb N, k\lt 81, k\ \text{is coprime to $9$})$ (note! not $123456789$) has every number from $0$ to $9$ except one number as its digit numbers .

$$12345679\times 2=[0]24691358$$ $$12345679\times 4=[0]49382716$$ $$12345679\times 5=[0]61728395$$ $$12345679\times 7=[0]86419753$$ $$12345679\times 8=[0]98765432$$ $$12345679\times 10=123456790$$ $$\vdots$$ $$12345679\times 77=950617283$$ $$12345679\times 79=975308641$$ $$12345679\times 80=987654320$$ $$(12345679\times 82=1012345678)$$

I've been thinking about its generalization.

Here is my question.

Question: Is the following proposition true?

Proposition: Let $n\ge2\in\mathbb Z.$ For any $k\in\mathbb N$ such that $k\lt n^2$ and $k$ is coprime to $n$, if we consider $$[1,2,3,\cdots, (n-3),(n-2),n]_{n+1}\times k$$ as a number with $n$-digits in base $n+1$, then it has every number from $0$ to $n$ except one number as its digit numbers, where $$[1,2,3,\cdots, (n-3),(n-2),n]_{n+1}$$ represents $123\cdots(n-3)(n-2)n$ in base $n+1$.

**Example** : The examples at the top are the $n=9$ cases of the proposition.

**Remark** : Suppose that the $n$-th digit number of a number with $n-1$ digits is $0$.

For example, suppose that we treat $$12345679\times 2=24691358$$ as $$12345679\times 2=[0]24691358.$$

**Motivation** : I've known the followings :

$$12345679\times 9=111111111$$ $$12345679\times 18=222222222$$ $$\vdots$$ $$12345679\times 72=888888888$$ $$12345679\times 81=999999999$$

Then, I found that the property at the top has already been known.

Though I've tried to prove that the proposition is true, I'm facing difficulty. Can anyone prove or disprove it?

**P.S. $1$** :

Before I posted this question at math overflow, but it was considered as off-topic. However, I got helpful comments from S. Carnahan there.

"Isn't this just a manifestation of the base $n+1$ expansion of $1/n^2$?"

"We may assume $k\lt n$, since the addition of $j/n$ just adds copies of $j$, preserving the pattern. The progression from one digit to the next in the expansion is addition by $k$ iff the next digit has no carry, and $k+1$ iff the next digit has a carry. The symbol $n−k$ is necessarily skipped, and there is no premature periodicity (since the multiplicative order of $n+1$ mod $n^2$ is unchanged). "

I think these comments must be true, but these are not obvious for me.

**P.S. $2$** :

I'm going to write the proof for the $k\lt n$ cases to get more hints from you. This is because I'm facing difficulty for proving the following though I think the proof for the $k\lt n$ cases would be fine.

What I cannot proveis : If the proposition is true for $k\lt n$, then it is true for $n\lt k\lt n^2.$

(Though this may be obvious, I cannot get a rigorous proof for it...please let me know it.)

In the following, I'm going to write **the proof for the $k\lt n$ cases**.

**Proof** : First, we use the following fact (of course, the proof is needed, but it is easy to prove it.)

Fact : $[1,2,3,\cdots,(n-3),(n-1),n]\times n=[1,1,1,\cdots, 1,1,1]$ (with $n$ $1$s)

Using this fact, we can see $$\begin{align}[1,2,3,\cdots, (n-3),(n-2),n]\times k&=[1,1,1,\cdots,1,1,1]\div n\times k\\&=[1,1,1,\cdots,1,1,1]\times k\div n\\&=[k,k,k,\cdots,k,k,k]\div n.\end{align}$$

In the following, let us consider in mod $n$.

We can see that the remainders in the process of calculating
$$[k,k,k,\cdots,k,k,k]\div n$$
are
$$k,2k,3k,\cdots,(n-1)k,nk(\equiv 0).$$
Since $k$ is coprime to $n$, we can easily see that there is no $i\not=j\ (1\le i,j\le n-1)$ such that
$$ik\equiv jk.$$
This immediately leads that the answer for
$$[k,k,k,\cdots,k,k,k]\div n$$
has distinct numbers as its digit numbers. **Q.E.D.**

**P.S. $3$** : Now I have a question.

Can I say the following at the last step of my proof above? I thought it was obvious, but now I feel this does not seem obvious...

"This immediately leads that the answer for $$[k,k,k,\cdots,k,k,k]\div n$$ has distinct numbers as its digit numbers."