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 $987654321$ $=$ $8$ $\times$ $123456789$ $+$ $9$ $98765432$ $=$ $8$ $\times$ $12345678$ $+$ $8$ $9876543$ $=$ $8$ $\times$ $1234567$ $+$ $7$ $987654$ $=$ $8$ $\times$ $123456$ $+$ $6$ ... $9$ $=$ $8$ $\times$ $1$ $+$ $1$

Saul Foresta explained as follows why this pattern holds in the decimal system and in other number systems using bases other than base $10$:

I generalized the problem for any base $n$ and any number of digits $r$ where $r$ can be anywhere from $1$ to $(n - 1)$.

Then after rewriting both sides of the equality given in the problem using sigma notation I arrived at the following:

$${\sum_{k=1}^r (n-k)n^{r-k}} = {{(n-2)\sum_{k=1}^r kn^{r-k} + r}}$$

In each summation $k$ stands for the $k$th digit of the number we're dealing with, reading from left to right. For example, in the number $9876$, $k$ ranges from $1$-$4$, where $9$ is $k=1$, $8$ is $k=2$, and so on.

So all I need to do in order to prove that this pattern holds is show that the left side of this equality does indeed equal the right side. Taking the terms like $8 \times123456789$, that is

$$(n-2)\sum_{k=1}^r kn^{r-k}$$

over to the left hand side, we will prove that this expression is equal to $r$.

[(n-1)n r-1 + (n-2)n r-2 + (n-3)n r-3 + ... + (n-r)] - (n-2)[n r-1 + 2n r-2 + 3n r-3 + ... + r] =
[n r - n r-1 + n r-1 - 2n r-2 + n r-2 - 3n r-3 + ...+ n - r] - [n r - 2n r-1 + 2n r-1 - 2.2n r-2 + ... + (n-2)r]

The coefficient of $n^{r-k}$ on this left hand side is $[1-k] - [k+1-2k] = 0$ for

$( 1 \le k \le r-1)$

and the coefficient of $n^r$ is also $0$.

The coefficient of $n^0$ is $[-r] - [-2(r)] = r$ and hence this expression is equal to $r$ as required.