For a given number k，the existence of prime p > k that satisfied 10, 100,...,10^(p-1) are complete residue system modulo p

Question

Is there a way to prove or disprove：

For any given number $k$，there must be prime $p$ that satisfied $10, 100,\dots,10^{p-1}$ are complete residue system modulo $p$.

for example： If k = 6, I can choose p = 7 or 17 10¹≡3,10²≡2,10³≡6,10⁴≡4,10⁵≡5,10⁶≡1 (mod 7) 10¹≡10,10²≡15,10³≡14,10⁴≡4,10⁵≡6,10⁶≡9,10⁷≡5,...10¹⁵≡8,10¹⁶≡12 (mod 17)

but if I choose p = 11 or 13 10¹≡10,10²≡1,10³≡10,10⁴≡1... (mod 11) 10¹≡10,10²≡9,10³≡12,10⁴≡3,10⁵≡4,10⁶≡1,10⁷≡10...(mod 13) then that is not a complete residue system

I know 7, 17,19, 23, 29, 47, 59 are the type of primes, and it seems related to Fermat little theorem? It is same to prove: for 1≦n≦(p-1), n satisfied 10ⁿ≡1 (mod p) is only (p-1), numbers of primes of the type are indefinitely.

Thank you very much for your help.

See https://oeis.org/A001913. The general case is not known: https://en.wikipedia.org/wiki/Artin%27s_conjecture_on_primitive_roots — lhf, Jun 22 '21 at 10:27
See also https://math.stackexchange.com/questions/2848716/10-is-a-primitive-root-modulo-p-4q1 — lhf, Jun 22 '21 at 10:38

score 1 · Accepted Answer · answered Jun 22 '21 at 21:14

No one knows whether the statement is true or false.

The statement is equivalent to the existence of infinitely many primes for which $10$ is a primitive root. This is an instance of Artin's conjecture on primitive roots. Not a single instance of this conjecture has been proved. There are some partial results.

For a given number k，the existence of prime p > k that satisfied 10, 100,...,10^(p-1) are complete residue system modulo p

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