Prove that for any natural number $n$, $1000^n-1$ is not a divisor of $1978^n-1$.

Question

This question was asked 7 years ago here and the answer has left me more stumped than I was before.

How is something like this proven?

Answer 1

The idea is that if $n$ is a multiple of $1, 3, 9, 27, ...$, respectively,

then $1000^n-1$ is a multiple of $27, 81, 243, 729, ...$, respectively,

but the powers of $3$ dividing $1978^n-1$ are only $3, 9, 27, 81, ..., $ respectively;

but if $1000^n-1$ were a divisor of $1978^n-1$,

then any factor of $1000^n-1$ would be a factor of $1978^n-1$.