Repunit 19 is Unique Period Prime with Period 19

Theorem
The decimal expansion of the reciprocal of the repunit prime $R_{19}$ has a period of $19$.
 * $\dfrac 1 {1 \, 111 \, 111 \, 111 \, 111 \, 111 \, 111} = 0 \cdotp \underbrace{\dot 000 \ldots 000 \ldots 000 \ldots 000 \ldots 000 \ldots 000} \dot 9$

This is the only prime number to have a period of exactly $19$.

Proof
The reciprocal of a repunit $R_n$ is of the form:
 * $\dfrac 1 {R_n} = 0 \cdotp \underbrace{\dot 000 \ldots 000}_{n - 1} \dot 9$

Thus $\dfrac 1 {R_{19}}$ has a period of $19$.

From Period of Reciprocal of Prime, for prime numbers such that:
 * $p \nmid 10$

we have that the period of such a prime is the order of $10$ modulo $p$.

That is, the smallest integer $d$ such that:
 * $10^d \equiv 1 \pmod p$

The only other possible primes $p$ whose reciprocals might have a period of $19$ must also satisfy:
 * $10^{19} \equiv 1 \pmod p$

that is:
 * $p \divides \paren {10^{19} - 1} = 9 \times R_{19}$

Therefore the only other possible prime whose reciprocal might have a period of $19$ is $3$.

Trivially:
 * $\dfrac 1 3 = 0 \cdotp \dot 3$

which has a period of $1$.

Hence the result.