= Solution
The <Fermat little theorem> states that if $p$ is a <prime number>, then $a^p\equiv a\pmod p$ for every <integer> $a$. Equivalently, for $p\nmid a$, $a^{p-1}\equiv1\pmod p$.
To prove it when $p\nmid a$, multiplication by $a$ permutes the nonzero residues $1,\ldots,p-1$. Indeed, $ai\equiv aj\pmod p$ implies $p\mid a(i-j)$, and the <Euclid lemma> implies $i=j$ in this range. Multiplying all these residues gives
$$
a^{p-1}(p-1)!\equiv(p-1)!\pmod p.
$$
None of the factors of $(p-1)!$ is divisible by $p$, so the <factorial> has a <modular inverse>. Cancelling it proves $a^{p-1}\equiv1$, hence $a^p\equiv a$. When $p\mid a$, the latter identity has both sides zero, completing the proof.
For an odd <prime number> $p\ne5$, the numbers $10,p$ are coprime. The <Fermat little theorem> gives $10^{p-1}\equiv1\pmod p$, and consequently
$$
\boxed{p\mid10^{j(p-1)}-1\quad\text{for every }j\ge1.}
$$
These exponents are distinct, so infinitely many are obtained.
Write the <repdigit> with $n$ copies of $5$ as
$$
R_n=5\sum_{j=0}^{n-1}10^j=\frac{5(10^n-1)}9.
$$
If $p\ne3,5$, choose $n$ to be any positive multiple of $p-1$. Then $9$ has a <modular inverse> modulo $p$, so $p\mid R_n$. Division by $9$ cannot be used modulo $3$; instead $10\equiv1\pmod3$ gives $R_n\equiv5n\pmod3$, so $3\mid R_{3j}$ for every $j\ge1$. If $p=5$, every $R_n$ is divisible by $5$. Thus
$$
\boxed{\text{Every odd prime divides infinitely many of }5,55,555,\ldots.}
$$
The separate treatment of $3$ accounts for the prime factor of the geometric-sum denominator.
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