Sum of product of digits

The sum of products of digits of 4-digit positive integers is given by:

$\begin{aligned} S & = \sum_{g=1}^9 \sum_{h=1}^9 \sum_{j=1}^9 \sum_{k=1}^9 ghjk = \sum_{g=1}^9 g \sum_{h=1}^9 h \sum_{j=1}^9 j \sum_{k=1}^9 k = 45^4 = \boxed{4100625} \end{aligned}$

Consider the expression $(1+2+3+...+9)(0+1+2+3+...+9)(0+1+2+3+...+9)(0+1+2+3+...+9)$ .

When expanded, it covers the product of every four digit number. Thus, our answer is $(1+2+3+...+9)^4=45^4=4100625$ .