We know that $10$ is a primitive root modulo $7^n$ for all $n$ (because its a primitive root for $7$ and for $7^2$ and we use the lemma). Now take $m$ such that $7^m>10^{2000}$. $10$ is a primitive root modulo $7^m$, so there is a number $s>m$ such that $10^s\equiv 7-6*10^m (mod 7^m)$. It implies that $\frac{10^m-1}{9}*7+\frac{10^s-10^m}{9}\equiv 0 (mod 7^m)$,(just do the equivalences and see its true) it means that $7^m$ divides $N=11...11177...77=\frac{10^m-1}{9}*7+\frac{10^s-10^m}{9}$. But the product of the digits of $N$ is $7^m$, so $N$ satisfies the condition, and the product of its digits is greater than $10^{2000}$

Shu wrote: Is there a positive integer divisible by the product of its digits such that this product is greater than $10^{2000}$? just let last n digits all be 1 or 2,when n is infinite,the number of "2" is infinite, so the product of its digits is infinite.so let $2^{n}$ divides the last n digits, Then the problem is solved.

horizon wrote: Shu wrote: Is there a positive integer divisible by the product of its digits such that this product is greater than $10^{2000}$? just let last n digits all be 1 or 2,when n is infinite,the number of "2" is infinite, so the product of its digits is infinite.so let $2^{n}$ divides the last n digits, Then the problem is solved. yes,this is a well-known lemma,and it can be recounted in the following form: for any $n$,there exists an n-digit integer which consists of only $1$ and $2$ and is divisible by $2^n$ it can be found even in Hongeegun's book 'Elementary Number Theory'.