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I guess you are talking about Fermat's Little Theorem, which was first published by (instead) Euler.

The theorem states that if prime P doesn't evenly divide A, then:

1.      A^(P-1) - 1 ≡ 0 (mod P)

No prime P : P > 5 evenly divides 10. (Primes 2 and 5 do, but they are obviously not > 5.) Therefore, for any prime P : P > 5 it is true that:

2.      10^(P-1) - 1 ≡ 0 (mod P)

Or, in other words, that all prime P : P > 5 must evenly divide at least one value in 99..9 form, namely 10^(P-1) - 1. We can restate (2) as:

3.      10^(P-1) ≡ 1 (mod P)

Now, you know that:

4.      a⋅b ≡(a mod c)⋅(b mod c) (mod c)

So, suppose:

5.      N = 10^(P-1) ⋅ 10^(P-1) = 10^(2⋅(P-1))

Then,

6.      N ≡ 10^(P-1) ⋅ 10^(P-1) (mod P)

7.      N ≡(10^(P-1) mod P)⋅(10^(P-1) mod P) (mod P)

8.      N ≡ 1⋅1 (mod P)

9.      N ≡ 1 (mod P)

Therefore,

10.     N-1 ≡ 0 (mod P)

Or P evenly divides N-1. But N=10^(2⋅(P-1)) and so N-1 is also a value in the 99..9 form, as well.

And you can repeat this process again by multiplying by another 10^(P-1) and using the above proof. So it must be that for all k : k ∈ ℕ, that:

11.     10^(k⋅(P-1)) - 1 ≡ 0 (mod P)

And since the set of ℕ is infinite, so must also be the set defined by (11) above.

Note that 99...9 [with n 9's] = 10^n - 1.

Fixing a prime p > 5, we have (since gcd(10, p) = 1) by Fermat's Little Theorem

that 10^(p-1) = 1 (mod p).

So, if we take n to be any positive multiple of p-1, we have 10^n = 1 (mod p).

Hence, any prime p > 5 divides infinitely many numbers of the form 99…9.

I hope this helps!