What will be $\text {Ord}_n\ q$? I am studying Galois theory from NPTEL lecture series on finite fields and Galois theory. While watching a lecture on cyclotomic polynomial I came across a theorem which I failed to understand partially. Here it is $:$
Theorem $:$
Let $p$ be a prime number and $q=p^e.$ Let $n \in \Bbb N$ be such that $p \nmid n$ and let $\Bbb F_q^{(n)}$ be the splitting of the polynomial $X^n-1$ over $\Bbb F_q.$ Let $[q]_n \in \Bbb Z_n^{\times}.$ Then $$\begin{align*} \left [\Bbb F_q^{(n)} : \Bbb F_q \right ] & = \#\ \text {Gal}\ \left (\Bbb F_q^{(n)} \bigg |\ \Bbb F_q \right ) \\ & = \text {Ord}_n\ q \\ & = \text {Ord}_n\ p^e \\ & = \frac {\text {Ord}_n\ p} {\text {gcd}\ (e,m)}. \end{align*}$$
Where $\text {Ord}_n\ x = \text {Ord}_{\Bbb Z_n^{\times}}\ [x]_n$ for any $[x] \in \Bbb Z_n^{\times}$ and $ \#\ \text {Gal}\ \left (\Bbb F_q^{(n)} \bigg |\ \Bbb F_q \right ) = m.$
I have almost understood the theorem except the last equality. I know that if $G$ be a finite group and $x \in G$ with $\text {Ord}\ (x)=n.$ Then $$\text{Ord}\ (x^m) = \frac {n} {\text {gcd}\ (m,n)}.$$
So in this case we should have $$\text {Ord}_n\ p^e = \frac {\text {Ord}_n\ p} {\text {gcd}\ (e,\text {Ord}_n\ p)}.$$
But this will imply that $$\text {gcd}\ (e,m) = \text {gcd}\ \left (e,\text {Ord}_n\ p \right ).$$
How do I show that? I observed that $$\begin{align*} {[q]_n}^{\text {Ord}_n\ p} & = {[p^e]_n}^{\text {Ord}_n\ p} \\ & = {[p]_n}^{\left (\text {Ord}_n\ p \right ).e} \\ & = \left ({[p]_n}^{\text {Ord}_n\ p} \right )^e \\ & = {[1]_n}^e \\ & = [1]_n. \end{align*}$$ This shows that $$m\ \big |\ \text {Ord}_n\ p\ \implies \text {gcd}\ (e,m)\ \bigg |\ \text {gcd}\ \left (e,\text {Ord}_n\ p \right ).$$
So if we can prove that $$\text {gcd}\ \left (e,\text {Ord}_n\ p \right )\ \bigg |\ \text {gcd}\ (e,m)$$ we are through. So we need only to show that ${[p]_n}^m=[1]_n.$ In other words $n\ \big |\ p^m-1.$ But how do I prove this?
Any help in this regard will be highly appreciated. Thank you very much for reading.
Source : https://youtu.be/A_andcVo1uU?list=PLOzRYVm0a65dsCb_gMYe3R-ZGs53jjw02&t=1176
 A: If ${[p]_n}^m = [1]_n$ then $\text {Ord}_n\ p = \text {Ord}_n\ q = m$ since $\text {Ord}_n\ q = m\ \bigg |\ \text {Ord}_n\ p.$ Now since $q = p^e$ so $\langle q \rangle \subseteq \langle p \rangle.$ But since $\#\ \langle q \rangle = \#\ \langle p \rangle$ so $\langle q \rangle = \langle p \rangle.$ Therefore $\exists$ $r \in \Bbb N$ such that $[p]_n={[q]_n}^r .$ So $[p]_n= {[p]_n}^{er}.$ Since $p \nmid n$ so $[p]_n \in {\Bbb Z_n}^{\times}.$ So multiplying both sides of the last equation by ${[p]_n}^{-1}$ we get ${[p]_n}^{er-1} = [1]_n.$ This implies that $m \mid er-1.$ So $\exists$ $k \in \Bbb Z$ such that $er-1= km.$ But that means $er-km = r e + (-k) m = 1.$ But this implies that $\text {gcd}\ (e,m) = 1.$ But that implies $\text {gcd}\ (e, \text {Ord}_n\ p) = 1$ since $\text {gcd}\ (e, \text {Ord}_n\ p) = \text {gcd}\ (e,m).$Which may not be true in general. 
For instance take $p = 5, e=3, n=7.$ Then what will happen? Then it is very easy to see that $m = \text {Ord}_7\ 5^3 = 2$ and $\text {Ord}_7\ 5 = 6.$ Hence $$\text {gcd}\ (e, \text {Ord}_n\ p) = \text {gcd}\ (3,\text {Ord}_7\ 5) = \text {gcd}\ (3,6) = 3 \neq 1.$$
In this case $$\#\ \text {Gal}\ \left ({\Bbb F_{125}}^{(7)}\ \bigg |\ \Bbb F_{125} \right ) = 2 \neq 6 = \frac {\text {Ord}_7\ 5} {\text {gcd}\ (3,2)}.$$
