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:

In each summation k stands for the kth 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 8x123456789, that is

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 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.