In the book of Frazer Jarvis on page 54 there is given a proposition about the discriminant . It is called
$\textbf{Proposition 3.31 }$ Suppose that $K=\mathbb{Q(\gamma)}$ , and that the minimal polynomial of $\gamma$ over $\mathbb{Q}$ is $f(x)\in \mathbb{Q(x)}$ of degree n . Then $$\Delta \lbrace 1,\gamma,\gamma^2,...,\gamma^{n-1} \rbrace=(-1)^{n(n-1)/2}N_{K/\mathbb{Q}}(f'(\gamma)) . $$
Proof. $ \Delta \lbrace 1,\gamma,\gamma^2,...,\gamma^{n-1} \rbrace=\prod_{i<j}(\gamma_i-\gamma_j)^2$
where the conjugates of $ \gamma$ are $\gamma_1 , \gamma_2 , ..., \gamma_n$ . Using that the conjugates of $\gamma$ are the roots of f(x) , and that it is monic we have $$ f(x)=\prod_{i=1}^n(x-\gamma_i)$$ It follows from the product rule that $$f'(x)=\sum_{k=1}^n \prod_{i \ne k}(x-\gamma_i) $$ and $$ f'(\gamma_j)=\prod_{i \ne j}(\gamma_j-\gamma_i)$$ Then come the things I do not understand
$N_{K/\mathbb{Q}}(f'(\gamma))=\prod_{j=1}^nf'(\gamma_j)=\prod_{j=1}^n\prod_{i \neq j} (\gamma_j-\gamma_i) $
$N_{K/\mathbb{Q}}(f'(\gamma))=\prod_{i<j}[-(\gamma_i-\gamma_j)^2]=(-1)^{n(n-1)/2}\Delta \lbrace 1,\gamma,\gamma^2,...,\gamma^{n-1} \rbrace$ .
From definition I know $$N_{K/\mathbb{Q}}(f'(\gamma))=\prod_{j=1}^n\sigma_j(f'(\gamma)) $$
Your penultimate like looks pretty clear to me.
In the last line, one is pairing together terms $\gamma_i-\gamma_j$ and $\gamma_j-\gamma_i$ from the double product on the previous line. They multiply to give $-(\gamma_i-\gamma_j)^2$. There's one such product for each pair $(i,j)$ with $1\le i<j\le n$. As there are $\frac12n(n-1)$ such pairs, the overall sign is $(-1)^{n(n-1)/2}$. The remaining factor $\prod_{i<j}(\gamma_i-\gamma_j)^2$ is the discriminant of $(1,\gamma,\gamma^2,\ldots)$ as noted at the start of the proof (basically the square of a Vandermonde determinant).