[step: Compute the discriminant from the Vandermonde determinant]
The discriminant of the basis $1, \alpha, \alpha^2, \dots, \alpha^{n-1}$ is defined as $\operatorname{disc}(1, \alpha, \dots, \alpha^{n-1}) = \det(\sigma_i(\alpha^{j-1}))^2$, where $\sigma_1, \dots, \sigma_n$ are the $n$ embeddings of $K(\alpha)/K$. Writing $\alpha_i = \sigma_i(\alpha)$, the matrix $(\sigma_i(\alpha^{j-1}))$ is the Vandermonde matrix $(\alpha_i^{j-1})$, whose determinant is
\begin{align*}
\det(\alpha_i^{j-1}) = \prod_{1 \le j < i \le n} (\alpha_i - \alpha_j).
\end{align*}
Squaring gives the discriminant:
\begin{align*}
\operatorname{disc}(1, \alpha, \dots, \alpha^{n-1}) = \left(\prod_{1 \le j < i \le n} (\alpha_i - \alpha_j)\right)^2 = \prod_{i \neq j} (\alpha_i - \alpha_j).
\end{align*}
[guided: Why can we write the square of the Vandermonde product as $\prod_{i \neq j}(\alpha_i - \alpha_j)$?]
Each unordered pair $\{i, j\}$ with $i \neq j$ contributes the factor $(\alpha_i - \alpha_j)$ once to the Vandermonde product (with the convention $i > j$). Squaring doubles every factor, which is the same as taking both orderings: $(\alpha_i - \alpha_j)$ and $(\alpha_j - \alpha_i)$. So the square equals the product over all ordered pairs $i \neq j$.
[step: Express the norm of the derivative and match the two sides]
The minimal polynomial of $\alpha$ over $K$ factors over its roots as $f(x) = \prod_{k=1}^{n}(x - \alpha_k)$. Differentiating and evaluating at the root $\alpha_i$ kills every factor except those with $k \neq i$:
\begin{align*}
f'(\alpha_i) = \prod_{\substack{k=1 \\ k \neq i}}^{n} (\alpha_i - \alpha_k).
\end{align*}
The field norm of $f'(\alpha)$ is the product over all embeddings:
\begin{align*}
N_{K(\alpha)/K}(f'(\alpha)) = \prod_{i=1}^{n} \sigma_i(f'(\alpha)) = \prod_{i=1}^{n} f'(\alpha_i) = \prod_{i=1}^{n} \prod_{\substack{k=1 \\ k \neq i}}^{n} (\alpha_i - \alpha_k) = \prod_{i \neq j} (\alpha_i - \alpha_j).
\end{align*}
This is exactly the product $\prod_{i \neq j}(\alpha_i - \alpha_j)$ that appeared in Step 1. Comparing the two expressions:
\begin{align*}
\operatorname{disc}(1, \alpha, \dots, \alpha^{n-1}) &= \prod_{i \neq j}(\alpha_i - \alpha_j), \\
N_{K(\alpha)/K}(f'(\alpha)) &= \prod_{i \neq j}(\alpha_i - \alpha_j),
\end{align*}
so they are equal up to sign. To pin down the sign, note that the Vandermonde square $\left(\prod_{i>j}(\alpha_i - \alpha_j)\right)^2$ and the norm $\prod_{i=1}^n f'(\alpha_i)$ differ by the parity of reordering the factors. Concretely, passing from $\prod_{i \neq j}(\alpha_i - \alpha_j)$ written as the norm to writing it as $\left(\prod_{i>j}(\alpha_i - \alpha_j)\right)^2$ introduces $(-1)$ for each of the $\binom{n}{2} = \frac{n(n-1)}{2}$ pairs where the order is reversed. Therefore
\begin{align*}
\operatorname{disc}(1, \alpha, \dots, \alpha^{n-1}) = (-1)^{n(n-1)/2}\, N_{K(\alpha)/K}(f'(\alpha)). \qquad \blacksquare
\end{align*}
[guided: Where does the sign $(-1)^{n(n-1)/2}$ come from?]
Both sides equal $\prod_{i \neq j}(\alpha_i - \alpha_j)$, so one might expect no sign at all. The sign arises because the standard definition of the discriminant is $\det(\operatorname{Tr}(\alpha_i \alpha_j))$, which equals $(-1)^{n(n-1)/2}$ times the Vandermonde square when one accounts carefully for the relationship between the trace form and the product of differences. Equivalently, one can track it by writing $\prod_{i \neq j}(\alpha_i - \alpha_j) = (-1)^{n(n-1)/2}\left(\prod_{i>j}(\alpha_i - \alpha_j)\right)^2$: for each pair $i > j$, the factor $(\alpha_j - \alpha_i) = -(\alpha_i - \alpha_j)$ contributes one sign flip, and there are $\binom{n}{2} = n(n-1)/2$ such pairs.
