I think the rest is indeed easy. Let m = 11...11 (p-1 ones). Suppose n = 11...11 ≡ 0 (mod p). Then 10^(p-1) * n + m ≡ 0 (mod p).
Prove that if p is prime and p > 5, then infinitely many members of the sequence 1, 11, 111, 1111, ... are divisible by p.
I think I can do most of it. Since p > 5, I used Fermat's little theorem to show 10^(p-1) - 1 ≡ 0 mod p. Then 99...99 ≡ 0 mod p (p-1 nines), which means 11...11 ≡ 0 mod p (p-1 ones) since p > 5 so gcd(p,9) = 1.
That shows one member of the sequence is divisible by p, but how do I go from here to show there are infinitely many?
Note that the numbers are exactly the numbers of the form , where is a positive integer.
As you've pointed out the member in the sequence is divisible by .
But what about any th member, where is a multiple of ?
We see that since divides and so . Lastly, since we may divide by to get .
It works for every prime not equal to or .