Let $R_n$ be the integer consisting of $n$ digits 1. That is,
$$R_n = \underbrace{111\ldots1}_{n\ \text{digits}} = \frac{10^n - 1}{9}.$$

Suppose $R_n$ is a prime number. We want to show that $n$ is also a prime number.

Assume, for contradiction, that $n$ is composite. Then $n = ab$ for integers $a, b > 1$.

Consider $R_a$:
$$R_a = \frac{10^a - 1}{9}.$$

Now, note that
$$10^n - 1 = 10^{ab} - 1 = (10^a)^b - 1 = (10^a - 1)\left(10^{a(b-1)} + 10^{a(b-2)} + \cdots + 10^a + 1\right).$$

Therefore,
$$R_n = \frac{10^n - 1}{9} = \frac{10^a - 1}{9} \cdot \left(10^{a(b-1)} + 10^{a(b-2)} + \cdots + 10^a + 1\right).$$

Since $a, b > 1$, both factors are greater than 1, so $R_n$ is composite, contradicting the assumption that $R_n$ is prime.

Therefore, if $R_n$ is prime, $n$ must be prime.