Proving $r^t \not\equiv 1\pmod p$

Question

$p>q$ are primes with $q \mid p-1$. Let $t=(p-1)/q$. I want to show the existence of $r$ such that $0<r<p$ and $r^t \not\equiv 1\pmod{p}$. It seems that there is no prime $p$ such that $p$ divides $2^t, 3^t, \cdots, (p-1)^t$ but why?

Answer 1

Let $r$ be a primitive root modulo $p$. Since $t<p-1$ and the order of $r$ is $p-1$, $r^t\not\equiv1\bmod p$. This implies the second statement is true, since $p$ was arbitrary.