If $p \equiv 1 \pmod{N!}$ prove there are no primitive roots $\pmod p$, $g$, that are less than $N$

Question

I've recently been looking at different questions and proofs in my book and one eludes me for the 2nd day in the row.

Let $N \geq 4$. Show that if $p$ is a prime such that $p \equiv 1 \pmod{N!}$ then none of the numbers $1, 2, \ldots, N$ are primitive roots modulo $p$.

I've tried using CRT, the fact that $\left(\frac{ab}{p}\right) = \left(\frac{a}{p}\right)\left(\frac{b}{p}\right)$ and Gaussian sum $G(p)$ properties, yet to no avail.

This proof has been really bugging me, could you please point me in the right direction?

Answer 1

We have $p \bmod N! = 1$ and $N > 3$ so we have $p \bmod 4 = 1$. Consider $q < N$ is a prime number we have $p \bmod q = 1$, so by quadratic reciprocity law we have $$\left(\frac{q}{p}\right) = \left(\frac{p}{q}\right) = \left(\frac{1}{q}\right) = 1$$ using $$\left(\frac{a}{p}\right)\left(\frac{b}{p}\right) = \left(\frac{ab}{p}\right)$$ we have $1, 2, \ldots, N$ are quadratic residue so they are not primitive roots.