Let $K=\mathbb{Q}(\alpha)$, $\alpha$ integer on $\mathbb{Q}.$ If $\alpha$ is a root of a p-Eisenstein polynomial, then p doesn’t divide the index
I want to prove what I have written in the title. So let $K=\mathbb{Q}(\alpha)$ the number field. I tried to prove this fact as follows: I know that $$ind(\alpha)=|\mathcal{O}_K/\mathbb{Z}[\alpha]|.$$ Let $p$ be the minimal polynomial of $\alpha:$ $p(x)=c_0+c_1x+\dots+c_{n-1}x^{n-1}+x^n,$ where the coefficients are integers, $p$ divide all of them and $p^2$ does not divide $c_0.$ Since $\alpha$ is an integer, I know that $$\mathbb{Z}[\alpha]=(1,\alpha,\alpha^2,\dots,\alpha^{n-1})_\mathbb{Z}$$ and there exists integers $a_i$ such that $$\mathcal{O}_K=(1,b_1\alpha,b_2\alpha^2,\dots,b_{n-1}\alpha^{n-1})_\mathbb{Z}$$ Then $$\mathcal{O}_K/\mathbb{Z}[\alpha]=\mathbb{Z}/b_1\mathbb{Z}\oplus\dots\oplus\mathbb{Z}/b_{n-1}\mathbb{Z},$$ $$ind(\alpha)=b_1\cdot\dots\cdot b_{n-1}$$ If $p$ divides the index, then it divides at least one of the $b_i.$ I would like to conclude using the fact that $b_i\alpha$ is an integer, but I don’t see how. Is my way a good way for proving this fact? How can I conclude?