1. Consider the expression $p^p - 1$. We can factorize it as:
$$p^p - 1 = (p-1)(p^{p-1} + p^{p-2} + \cdots + p + 1)$$
Notice that $p^{p-1} + p^{p-2} + \cdots + p + 1$ is a sum of $p$ terms, each of which is greater than or equal to 1. Therefore, $p^{p-1} + p^{p-2} + \cdots + p + 1 > 1$.

2. Since $p^{p-1} + p^{p-2} + \cdots + p + 1$ is greater than 1, it must have a prime factor $q$. Thus, we have:
$$q \mid p^{p-1} + p^{p-2} + \cdots + p + 1$$
and consequently,
$$q \mid p^p - 1$$
This implies:
$$p^p \equiv 1 \pmod{q}$$

3. By definition, the order of $p$ modulo $q$, denoted as $\text{ord}_q(p)$, is the smallest positive integer $k$ such that:
$$p^k \equiv 1 \pmod{q}$$
Since $p^p \equiv 1 \pmod{q}$, it follows that $\text{ord}_q(p)$ divides $p$. Given that $p$ is a prime number, the possible values for $\text{ord}_q(p)$ are 1 or $p$.

4. If $\text{ord}_q(p) = 1$, then:
$$p \equiv 1 \pmod{q}$$
Substituting $p \equiv 1 \pmod{q}$ into the sum $p^{p-1} + p^{p-2} + \cdots + p + 1$, we get:
$$1^{p-1} + 1^{p-2} + \cdots + 1 + 1 = p \equiv 1 \pmod{q}$$
This implies:
$$q \mid p^{p-1} + p^{p-2} + \cdots + p + 1$$
which contradicts the fact that $q$ is a prime factor of $p^{p-1} + p^{p-2} + \cdots + p + 1$. Therefore, $\text{ord}_q(p) = 1$ is not possible.

5. Hence, we must have $\text{ord}_q(p) = p$. Since $\text{ord}_q(p)$ divides $q-1$ (by Fermat's Little Theorem), it follows that:
$$p \mid q-1$$
Therefore:
$$q \equiv 1 \pmod{p}$$

6. Since $q$ is a prime factor of $p^p - 1$ and $q \equiv 1 \pmod{p}$, we have shown that $p^p - 1$ has a prime factor that is congruent to 1 modulo $p$.

$\blacksquare$