$K|\mathbb{Q}_p$ un-ramified if and only if $(d_K)=(1)$: help with a passage I need some help in th last passage of this proof:

Suppose $K|\mathbb{Q}_p$ is un-ramified and of degree $n$. then $K=\mathbb{Q}_p(\alpha)$, where $\alpha$ can be taken to be an integral unit in $\mathcal{O}_K$. It's enough to show that the discriminant of $K|\mathbb{Q}_p$, denoted with $d_K$, is the trivial ideal $(1)$. We can use the integral basis $\{\alpha^i\}_1^n$ to compute it, and the fact that the extensions is separable. So $$(d_k)=(\det (\sigma_j\alpha^i))^2$$ where $\sigma_j$ are the $n$ embeddings of $K$ over $\mathbb{Q}_p$ in the algebraic closure of $K$. It's easy to see that such determinant is a Vandermonde determinant and it's equal to $$ \prod_{j<k}(\sigma_j\alpha - \sigma_k \alpha)^2$$

Then according to my lecturer, such determinant is clearly an integral unit, ending the proof. I really don't see why this is true: I tried to prove that each single element in the product is invertible (in the integral ring $\mathbb{Z}_p$) but the presence of the minus sign ruins this approach. I fear there is some kind of trick in this case, but I can't see it. Any suggestion?
 A: First of all, it seems that you are trying to prove the implication "$K/\mathbb Q_p$ unramified $\Rightarrow$ $\mathfrak d_{K/\mathbb Q_p} = \mathcal O_K$". Also, you didn't tell us what definition of unramified you are using (this might be relevant when you try to prove that two definitions are equivalent...). I will use:
Lemma. Let $K/\mathbb Q_p$ be a finite extension. Then $K$ is unramified if and only if the reduction map
\begin{align*}
f \colon \operatorname{Gal}(K/\mathbb Q_p) &\to \operatorname{Gal}(k/\mathbb F_p)\\
\sigma &\mapsto \bar \sigma
\end{align*}
is an isomorphism.
Answer to your question. If $K/\mathbb Q_p$ is unramified, then $f$ is an isomorphism. In particular, if $\sigma_i \neq \sigma_j$, then $\bar \sigma_i \neq \bar \sigma_j$. This shows that $\sigma_i(\alpha) - \sigma_j(\alpha)$ has to be a unit: if not, then $\bar \sigma_i(\bar \alpha) = \bar \sigma_j(\bar \alpha)$ (since $\mathbb Z_p$ is a local ring), which forces $\bar \sigma_i = \bar \sigma_j$ (since $\bar \alpha$ generates $k$ over $\mathbb F_p$), contradiction. $\square$
