Summation of products

Let $a(n)$ be the product of digits of number $n$ $\sum_{n=100}^{999} a(n)=\sum_{i=1}^{9} \sum_{j=0}^{9} \sum_{k=0}^{9} (ijk)=(\sum_{i=1}^{9} i)(\sum_{j=0}^{9} j)(\sum_{k=0}^{9} k)=(\sum_{i=1}^{9} i)^3=45^3$

More generally,let $S_n$ denote the answer when we are dealing with $n$ -digit numbers.

We shall prove by induction that $S_n=45^n$

For $n=1$ , $S_1=1+2+3+4+5+6+7+8+9=45$

Suppose the result holds for some $n\geq 1$ Consider now $(n+1)$ digit numbers whose first digit is $k$ where $k$ is any one of $1,2,...,9$ .

The sum of the products of the digits of these numbers is equal to the common factor $k$ times $S_n$

$S_{n+1}=(1+2+3+4+5+6+7+8+9)S_n=45^{n+1}$

This completes the inductive argument.In particular, $S_3=\boxed{45^3=91125}$