What is the smallest multiple of $3^{1000}$, which has only 1’s in the digits?(Like 1, 11, 111 etc) I think the answer is $3^{1000}$ ones,
and I want to prove it with induction. 
$3^0 | 1$ and $1$ is the smallest.
$ 3^1 | 111 $; it’s the smallest too. 
Let’s say it's true for $ 3^k | 111...111 $  ($3^k$  ones).   How can I prove to $3^{k+1} $ ?

If $m$ is coprime to $3$ then the g.c.d. of $10^m-1$ and $10^3-1$ is $10-1=9$. In particular, $9$ but not $27$ is  a factor of $10^m-1$.
If $n=3^km$ where  $m$ is coprime to $3$ then $$\frac {10^{3^km}-1}{10^{3^{k-1}m}-1}=1+10^{3^{k-1}m}+10^{3^{k-1}(2m)}\equiv3\pmod9$$
Repeating this process we obtain 
$10^{3^km}-1=\frac {10^{3^km}-1}{10^{3^{k-1}m}-1}\frac {10^{3^{k-1}m}-1}{10^{3^{k-2}m}-1}..\frac {10^{3m}-1}{10^{m}-1}(10^{m}-1)$
$$\equiv3^{k+2}\pmod {3^{k+3}}$$
Hence the least $n$ such that  $10^n-1$ is a multiple of $3^{1002}$ is obtained when $k=1000$ and $m=1$ i.e. $n=3^{1000}$.
