Let $K \subseteq L \subseteq M$ be finite extensions of number fields. Consider the following two formulas:
1) $\mathrm{Disc}_{M \mid K} = \mathrm{N}_{L \mid K}(\mathrm{Disc}_{M \mid L}) \cdot \mathrm{Disc}_{L \mid K}^{[M:L]}$.
Here, $\mathrm{Disc}_{M \mid K} = \mathrm{N}_{M \mid K}(\delta_{M \mid K})$, where $\delta_{M \mid K} = (\delta_{M \mid K}^{-1})^{-1}$ and $\delta_{M \mid K}^{-1} = \{ x \in M \mid \mathrm{Tr}_{M \mid K}(xy) \in \mathcal{O}_K \text{ for every } y \in \mathcal{O}_M \}$. Further, we define $\mathrm{N}_{L \mid K}(\mathfrak{P}) = \mathfrak{p}^{f(\mathfrak{P} \mid \mathfrak{p})}$ for every prime ideal $\mathfrak{P}$ in $L$ so that $\mathfrak{p} = \mathfrak{P} \cap \mathcal{O}_K$, and extend this definition multiplicatively to any fractional ideal in $L$.
2) Let $\{\alpha_1,...,\alpha_n\}$ be a basis for $L \mid K$ and $\{\beta_1,...,\beta_m\}$ be a basis for $M \mid L$, i.e. $n = [L:K]$ and $m = [M:L]$. Then $\{\alpha_1 \beta_1, ..., \alpha_n \beta_m\}$ is a basis for $M \mid K$ and $$\mathrm{disc}_{M \mid K}(\alpha_1 \beta_1, ..., \alpha_n \beta_m) = (\mathrm{disc}_{L \mid K}(\alpha_1,...,\alpha_n))^m \cdot \mathrm{N}_{L \mid K}(\mathrm{disc}_{M \mid L}(\beta_1,...,\beta_m)).$$ Here, we define as usually $\mathrm{disc}_{L \mid K}(\alpha_1,...,\alpha_n) = \det \big( (\mathrm{Tr}_{L \mid K}(\alpha_i \alpha_j))_{i,j} \big)$.
My question: Is it possible to derive formula 2) from formula 1)? (Maybe at least in the special case $K= \mathbb{Q}$, where simply $\mathrm{Disc}_{M \mid \mathbb{Q}} = (\Delta_M)$?) I am grateful for any help.