No, because for an $n$ digit number the digit sum is $\le 9n$ and the digit product is $\le 9^n$. The product of the two is therefore $\le 9^{n+1}n$.
An $n$ digit number is $\ge 10^{n-1}$
You would therefore need $81n\cdot9^{n-1}\gt 10^{n-1}$ or $81n\ge \left(\frac {10}{9}\right)^{n-1}$.
This is true for only finitely many values of $n$.