Let $F$ be a field with characteristic $0$ and $f\in K[X]$ not constant. Furthermore, let $L$ be an extension of $K$ such that $$f(X)=c(X-\alpha_1)^{k_1}\cdots(X-\alpha_n)^{k_n}$$ where $c,\alpha_1,\dots,\alpha_n\in L$, $k_1,\dots,k_n\in\mathbb N$ and $\alpha_i\neq\alpha_j$ for $i\neq j$.
Now one can show that $$q(X):=(X-\alpha_1)\cdots(X-\alpha_n)\in K[X]$$ and I am interested in a proof of that. I know one proof using the formal derivation on polynomial rings to argue that $(X-\alpha_i)^{k_i-1}$ is a common factor of $f$ and $f'$ for all $i\in\{1,\dots,n\}$ so that $$\gcd(f,f')=\prod_{i=1}^n(X-\alpha_i)^{k_i-1}$$
Is there a another way of obtaining this result?
An idea for you:
First, we assume $\;\alpha_1,\ldots, \alpha_n\notin K\;$ , otherwise we can remove the correspondent factor: if for example $\;\alpha_1\in K\;$, then
$$\;q(x)\in K[x]\iff (x-\alpha_2)\cdot\ldots\cdot(x-\alpha_n)\in K[x]$$
Now, let $\;h_1(x)\in K[x]\;$ be the minimal (monic, of course) polynomial of $\;\alpha_1\;$ over $\;K\;$. Since also $\;f(\alpha_1)=0\;$ and $\;f(x)\in K[x]\;$ , we get that $\;h_1\,\mid\,f\;$ , so $\;f(x)=h_1(x)g(x)\;$ , for some $\;g(x)\in K[x]\;$ .
Suppose that out of all the different roots of $\;f\;$ , namely $\;\alpha_1,...,\alpha_n\;$, the roots $\;a_1,...,a_{r_1}\;$ (if needed, we can renumber those different roots so the first $\;r_1\;$ are the roots of $\;h_1$) are roots of $\;h_1\;$ (since $\;h_1\;$ cannot have multiple (or repeated) roots...why?) .
Let now $\;h_2(x)\in K[x]\;$ be the minimal polynomial of $\;\alpha_{r_1+1}\;$ over $\;K\;$ . As before, let $\;\alpha_{r_1+1},...,\alpha_{r_2}\;$ be all the roots of $\;h_2\;$ out of the different roots of $\;f\;$.
Continue with the above inductive procedure, so that in the end we get polynomials $\;h_1,h_2,...,h_m\;$ which are minimal polynomials over $\;K\;$ of the different roots of $\;f\;$ .
Finally, observe that $\;h_1\cdot\ldots\cdot h_m=q(x)\in K[x]\;$