Proof that p Divides the Repunit R_p
The prime
\[p\]
divides the pth repunit \[R_{p-1}=\underbrace{11111...1111}_{(p-1) \; 1's}\]
Proof
We can use the identity
\[R_{p-1}=\frac{10^{p-1}-1}{9}\]
and Fermat's Little Theorem \[a^{p-1} \equiv 1 \; (mod \; p)\]
.\[9R_{p-1}=10^{p-1}-1 \equiv 0 \; (mod \; p)\]
by rearranging the identity the using Fermat's Little Theorem.Hence
\[p\]
divides \[9\]
or \[p\]
divides \[R_{p-1}\]
. The first option is impossible for \[p \gt 3\]
so in this case \[p\]
divides \[R_{p-1}\]
.If
\[p=3\]
then \[3\]
divides \[111=R_3\]
. 
