Let $b$ be the base of numeration.

Suppose that an integer $n$ has $m$ digits when expressed in base $b$ (not counting leading zeros, of course). Then $n \ge b^{m-1}$ .

Since each digit is at most $b-1$ , we have that the sum of the digits is at most $m(b-1)$ and the product is at most $(b-1)^m$ , hence the sum of the digits of $n$ times the product of the digits of $n$ is at most $m(b-1)^{m+1}$ .

If $n$ is a sum-product number, then $n$ equals the sum of its digits times the product of its digits. In light of the inequalities of the last two paragraphs, this implies that $m(b-1)^{m+1} \ge n \ge b^{m-1}$ , so $m(b-1)^{m+1} \ge b^{m-1}$ . Dividing both sides, we obtain $(b-1)^2 m \ge (b/(b-1))^{m-1}$ . By the growth of exponential function, there can only be a finite number of values of $m$ for which this is true. Hence, there is a finite limit to the number of digits of $n$ , so there can only be a finite number of sum-product numbers to any given base $b$ .