Repunit Integer as Product of Base - 1 by Increasing Digit Integer

Theorem
That is:
 * $\ds 9 \sum_{j \mathop = 0}^n \paren {n - j} 10^j + n + 1 = \sum_{j \mathop = 0}^n 10^j$

Proof
A specific instance of the general result:

where $b = 10$.