We know that if $\gcd(n,p)=1$, then the polynomial $x^n-1$ can be factored to the irreducible polynomials over $GF(p)$ by cyclotomic cosets method, as follows($t$ is a number of cyclotomic cosets) $$ x^n-1=f_1(x)\, f_2(x)\, \cdots \, f_t(x) $$
Now, we want to see the question of factoring $x^n-1=0$ over $GF(p)$ with another point of view. Let $k$ be the first positive integer such that $n\mid p^k-1$, which means $GF(p^k)$ have elements of the order $n$. Suppose that $\beta$
be an element of $GF(p^k)$ with order $n$, then $x^n-1$ can be represent as
$$
x^n-1=(x-1)(x-\beta)(x-\beta^2)\cdots (x-\beta^{n-1})
$$
My question: How to prove that the following relation( without considering cyclotomic cosets method) $$ (x-1)(x-\beta)(x-\beta^2)\cdots (x-\beta^{n-1})=f_1(x)\, f_2(x)\, \cdots \, f_t(x) $$
Example: By cyclotomic cosets method, $x^{13}-1$ over $GF(3)$ is factored as follows $$ x^{13}-1=(2+x)(2+x+x^2+x^3)(2+x^2+x^3)(2+2x+2x^2+x^3)(2+2x+x^3) $$
Now, we have $13\mid 3^3-1$ which results that if $\beta$ be an element of order $13$ over $GF(3^3)$that is constructed by $x^3+2x+1$, then we have $$ x^{13}-1=(x-1)(x-\beta)(x-\beta^2)\cdots (x-\beta^{12}) $$
Thanks for any suggestions.