Let $q$ be a prime number. Show that any prime divisor $p$ of $1+a+a^2+...+a^{q-1}$ satisfies $p \equiv1(q)$ or $p=q.$ Deduce that there are infinitely many primes $p\equiv1(q).$ The solution says: Notice that $$1+a+a^2+...+a^{q-1}=\frac{a^q-1}{a-1}, $$ if $a \neq1.$ Hence if $$1+a+a^2+...+a^{q-1}\equiv0(p),$$ then either $a^q\equiv1(p)$ and $a \not\equiv1(p)$ or $a\equiv1(p).$ In the former case $q|p-1,$ since $a$ has order $q$. Notice that any prime divisor of $2^q-1$ is congruent to $1 \ (mod\ q).$ Thus, there is at least one prime congruent to $1 \ (mod\ q).$ If there are only finitely many such primes, let us list them as $$p_1,p_2,...,p_k.$$ Then, putting $a=qp_1\cdots p_k,$ we find that any prime divisor $p$ of $$1+a+a^2+...+a^{q-1}$$ is first, coprime to $a=qp_1\cdots p_k,$ and second, must be congruent to $1 \ (mod\ q)$ or equal to $q,$ which is a contradiction.
In this proof I do not understand the part "... and second...".Why does $p=q$ follow ? If we take a prime number $p$ that is not in the list and set $a=qp_1\cdots p_k+1,$ then we know $a\equiv1(p)$ leading to $p|qp_1\cdots p_k$. So we have $p=q.$
What we know is that every prime divisor p of $1 + a + ... + a^{q-1}$ has to be congruent to $1 (mod q)$ or equal to q. Now if we let $a=qp_1...p_k$, then if the prime divisor is $q$, $a=qp_1...p_k$ and q have a common divisor. If the prime divisor is congruent to $1(mod q)$ it must be in $p_1 , ... , p_k$, since these were by assumption all prime numbers congruent to 1. In both cases $a$ and the primedivisor of $1 + a + ... + a^{q-1}$ have a common divisor, which cannot be true.