As the person from Math Overflow hinted, your question is best viewed in the paradigm of the base $n$ expansion of $1/(n-1)^2$.
Consider the following sum: $\displaystyle \sum_{k=1}^\infty \frac{k}{n^{k+1}}$. This is equal to $\displaystyle \frac{1}{n^2} + \frac{2}{n^3} + \frac{3}{n^4} + \frac{4}{n^5} + \cdots$ onward to infinity, which is in turn just equal to $\displaystyle \frac{n}{(n-1)^2}$.
The number $1n^{n-3} + 2n^{n-4} + 3n^{n-5} + \cdots + (n-3)n^1 + (n-1)$) (which is $12345679$ in base $10$) is simply this expansion multiplied by $n^{n-2}$, to produce $\displaystyle \frac{n^{n-1}}{(n-1)^2}$, and then truncating the fractional portion. So, since we're just shifting decimal places, let's look at what happens in just the case of $\displaystyle \frac{1}{(n-1)^2}$ because it's easier to work with.
In the case of $\displaystyle \frac{1}{(n-1)^2}$, the expansion in base $n$ is equal to $$ \frac{1}{n^2} + \frac{2}{n^3} + \frac{3}{n^4} + \frac{4}{n^5} + \cdots + \frac{n-1}{n^{n}} + \frac{n}{n^{n+1}} + \cdots$$
Now, here's an example of that in action in base 10:
0.0 1 2 3 4 5 6 7 9 0 1 2 3 ...
0 4 8 1 2 ...
1 5 9 1 3 ...
2 6 1 0 ...
3 7 1 1 ...
You'll notice that the tens digit of 10 butts into the place value of the ones digit of 9, causing a carry over and increasing the value of 8 in that place to 9.
For multiples of $\displaystyle \frac{1}{(n-1)^2}$ = $\displaystyle \frac{k}{(n-1)^2}$ , the same deal applies only in multiples. For $k = 2$, for example, $\displaystyle \frac{2}{(n-1)^2}$ is just equal to:
$$ \frac{2}{n^2} + \frac{4}{n^3} + \frac{6}{n^4} + \frac{8}{n^5} + \cdots + \frac{2(n-1)}{n^{n}} + \frac{2n}{n^{n+1}} + \cdots$$
All that changes is that the numbers being added as place values go into two digits more quickly.
0.0 2 4 6 9 1 3 5 8 0 2 4 6 ...
0 8 1 6 2 4 ...
2 1 0 1 8 2 6 ...
4 1 2 2 0 ...
6 1 4 2 2 ...
Notice that the digit increases by $2$ each time now, except when the tens digit of the next number increases by $1$, in which case it increases by $3$.
Also notice that there is no $7 (= 9 - 2)$ in this expansion, just as there was no $8(= 9 - 1)$ in the first one. This is because when you try to get $7$ (e.g. with the $6$ in $16$ + the $1$ in $18$), the $2$ in $20$ butts in and forces the $7$ to carry again, increasing it to $8$.
This basically means that whenever the ones digit would become a number that's $7$ or greater, it increases by $3$ rather than $2$. But that just means that we can treat $7$ as skipped entirely, and in fact we're actually always just moving $2$ steps forward in the cyclic sequence $0, 1, 2, 3, 4, 5, 6, 8, 9$.
And this applies in general, for $k/(n-1^2)$, we're moving $k$ steps forward in the sequence $[0, 1, 2, 3, 4, 5, 6, \cdots, n-1]$ with $(n - 1 - k)$ removed. This "trick" even works with numbers that aren't relatively prime to $n - 1$:
$$ 3 / 81 = 0.037037037037037... = \text{3 steps forward in } [0, 1, 2, 3, 4, 5, 7, 8, 9] $$
$$ 9 / 81 = 0.111111111111111... = \text{9 steps forward in } [1, 2, 3, 4, 5, 6, 7, 8, 9] $$
In base 7:
$$ 2_7 / 51_7 = 0.025025025025025..._7 = \text{2 steps forward in } [0, 1, 2, 3, 5, 6] $$
$$ 3_7 / 51_7 = 0.040404040404040..._7 = \text{3 steps forward in } [0, 1, 2, 4,
5, 6] $$
Keep in mind, though, that as soon as you get through the first set of $n - 1$, you go back to one step but everything in the original number sequence shifts to the right by one:
$$ 9 / 81 = 0.111111111111111... = \text{9 steps forward in } [1, 2, 3, 4, 5, 6, 7, 8, 9] $$
$$ 10 / 81 = 0.123456790123456... = \text{1 step forward in } [1, 2, 3, 4, 5, 6, 7, 9, 0] $$
$$ 18 / 81 = 0.222222222222222... = \text{9 steps forward in } [2, 3, 4, 5, 6, 7, 8, 9, 1] $$
$$ 19 / 81 = 0.234567901234567... = \text{1 step forward in } [2, 3, 4, 5, 6, 7, 9, 0, 1] $$
So the fact that all the digits except for one we removed are represented is simply a consequence of the fact that the order of $k$ in the cyclic group $\mathbb{Z}_{n-1}$ is still $(n-1)$ when $k$ is relatively prime to $(n-1)$.
This doesn't quite constitute a rigorous proof (mostly because I'm too lazy to formalize it and it took quite a while for me to wrap my head around what I was saying myself), but I think you have enough of an idea now that you can figure it out yourself.
As an additional note, it is in fact true that the multiples of $123456789$ (note! not $12345679$ which I explained above) that are relatively prime to $9$ also result in a permutation of those digits (including 0 when you reach 10 digits):
$$1 \times 123456789 = 123456789$$
$$2 \times 123456789 = 246913578$$
$$4 \times 123456789 = 493827156$$
$$5 \times 123456789 = 617283945$$
$$7 \times 123456789 = 864197523$$
$$8 \times 123456789 = 987654312$$
$$10 \times 123456789 = 1234567890$$
$$etc.$$
Well, up until 19, anyway:
$$19 \times 123456789 = 2345678991$$
You also asked for a proof that if the proposition holds for $k < n$, it also holds for $n \le k \le n^2$. This is actually just a simple induction, if you've already proven the $k < n$ cases.
Suppose that we have a sequence of digits $a b c d \ldots$ that contains all distinct digits except for $n - k$ in the sequence described above (i.e. the digit increases by $k$ each time now, except when it would exceed $n - k$, in which case it increases by $k+1$.).
Then increasing each digit by $1$ (as adding $n$ to the numerator will do) will mean that all the digits are still distinct, except for one that carries over to zero, increasing the previous digit by $1$. But the one immediately before it became $n - k$ upon increasing by 1, so increasing it to $n - k + 1$ means that it's still not the same as any other digits (because the one that was $n - k + 1$ before has since increased to $n - k + 2$). And the resulting sequence also has the property that "the digit increases by $k$ each time now, except when it would exceed $n - k$, in which case it increases by $k+1$", so the induction can continue going on.
Best Answer
I can prove a very narrow form.
Let's consider numbers of the form:
$(D_{n-1}D_{n-2}.....D_{2}D_{1}D_0)_{10} = (D_{n-1}D_{n-2}.....D_2[A..F])_{16}$.
Here 2 least significant digits in decimal representation change to [A..F]. For these numbers, conditions that need to satisfy are,
$100*x + 10 = n$ ....(1)
$16*y + 10 = n$ ....(2)
So,
$y = 6.25*x$ ....(3)
Let's assume,
$x_{10}$ is of form $....N_{k-1}N_{k-2}.....N_{2}N_{1}N_{0}$
or,
$x_{10} = ....+ (N_{k-1}10^{k-1}) + (N_{k-2}10^{k-2}) + ....+ (N_{2}10^2) + (N_{1}10^1) + N_{0}$
So,
$y_{10} = ....+ (N_{k-1}16^{k-1}) + (N_{k-2}16^{k-2}) + ....+ (N_{2}16^2) + (N_{1}16^1) + N_{0}$
Also from (3),
$y_{10} = 6.25(....+ (N_{k-1}10^{k-1}) + (N_{k-2}10^{k-2}) + ....+ (N_{2}10^2) + (N_{1}10^1) + N_{0})$
So,
$(....+ (N_{k-1}16^{k-1}) + (N_{k-2}16^{k-2}) + ....+ (N_{2}16^2) + (N_{1}16) + N_{0}) = 6.25(....+ (N_{k-1}10^{k-1}) + (N_{k-2}10^{k-2}) + ....+ (N_{2}10^2) + (N_{1}10^1) + N_{0})$
Using up to 6 digits for x,
$(1048576N_5 + 65536N_4 + 4096N_3 + 256N_2 + 16N_1 + N_0) = (625000N_5 + 62500N_4 + 6250N_3 + 625N_2 + 62.5N_1 + 6.25N_0)$
or,
$42357600N_5 + 303600N_4 - 215400N_3 - 36900N_2 - 4650N_1 - 525N_0 = 0$
Value in Hex falls behind till N3 because initial modulus was 100 for decimal and 16 for hex. But at and after N4, hex value overtakes decimal forever.
Only solution for N5 (and above) between 0 and 9 is 0. Also, there are no solutions possible less than 5 digits for x in (1).
So essentially numbers of these forms are only 7-digits or 2 digits, And only possible solutions are,