I am trying to proof that the first reappearing remainder when dividing one by a prime number is one.
What I found is that if the expansion of $1/p$ recurs with period $k$ then $10^k-1$ is divisible by $p$.
What I don't see is how this relates to my question although it somehow should, I guess.
Thank you!
EDIT
Is it perhaps possible to proof that the first reappearing remainder when dividing one by a prime number is one directly follows from the condition that if the expansion of $1/p$ recurs with period $k$ then $10^k-1$ is divisible by $p$?