Show that there is a positive integer $m$ such that $\prod_{σ∈Gal(L|F)}(x − σ(α)) = h^m(x)$

Let $$L$$ be a finite field extension of $$F$$ and let $$α ∈ L$$. Also, let $$α_1 = α, α_2, · · · , α_r$$ be the distinct elements of $$L$$ obtained by applying the elements of $$Gal(L|F)$$ to $$α$$. Let us consider the polynomial $$h(x) = \prod^r_ {j=1} (x − α_j ) ∈ L[x]$$. I wish to show that there is a positive integer $$m$$ such that $$\prod_{σ∈Gal(L|F)}(x − σ(α)) = h^m(x)$$

Not sure if it's relevant but I know from a textbook proof of "$$F$$ is the fixed field of $$Gal (L/F)$$ acting on $$L$$ implies $$F ⊂ L$$ is a normal separable extension" that this $$h ∈ F[x]$$ and that $$h$$ is irreducible over $$F$$, thus is the minimal polynomial of $$α$$ over F. The above formula for $$h$$ also shows that $$h$$ is separable and splits completely over L, but I don't know how to go about proving the desired result still hmm~

• This is basically restatement of the following. Denote $K$ as splitting field of $a$ over $F$. Now $Gal(L/K)\times Gal(K/F)\to Gal(L/F)$ is bijection and the morphism is obvious by composition. So every automorphism of $K/F$ extends $|Gal(L/K)|$ different ways. In particular, you see $a_1=\sigma(a)$ showing up exactly $|Gal(L/K)|$ times. This holds for any $a_i$ roots. So $m=[L:K]$ where $K$ is the splitting field of $a$ over $F$. Oct 6 '18 at 17:56
• Ohh does m relate to r? Oct 6 '18 at 18:33

Since the action of $$Gal(L/F)$$ on $$\{\alpha_1,\dots,\alpha_r\}$$ is a group action, each of the $$(x-\alpha_i)$$ will appear the same number of times in the product.