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?