Prove Hilbert's 90 in characteristic $0$ using discriminant I 'm reading A Brief Guide to Algebraic Number Theory by Swinnerton-Dyer. He proves Hilbert's 90 at page 4~5 in lemma 4(assuming characteristic $0$):

lemma 4 Let $K/k$ be a Galois extension whose Galois group
$Gal(K/k)$
is cyclic with generator $\sigma$. If $\alpha$ in $K$ is such that
$\mathrm{norm}_{K/k} \alpha =1$ then $\alpha=\beta/\sigma \beta$ for
some $\beta$ in $K$; and we can take $\beta$ to be integral.
Proof Let $[K:k]=n$ and for any $\gamma$ in $K$ consider  $$\beta=\gamma \cdot \alpha + \sigma \gamma \cdot \alpha \cdot \sigma 
 \alpha +\cdots + (\sigma^{n-1}\gamma)(\alpha \cdots 
 \sigma^{n-1}\alpha) $$        then $\alpha \cdot \sigma \beta =
\beta$. If $\beta=0$ for every  $\gamma$ then
$\Delta^2_{K/k}(\gamma_1,...,\gamma_n)=0$ for any
$\gamma_1,...,\gamma_n$ in $K$, and this we know to be false.

I can't see why $\beta=0$ for every $\gamma$ implies the discirminant is zero, could you give me some help? Thank in advance.

Hint: There is a formula for the discriminant in terms of $\sigma^i(\gamma_j)$, namely
$$
\Delta(\gamma_1,\ldots ,\gamma_n)=\det ((\sigma^i(\gamma_j))_{i,j})^2.
$$
