The case $\sigma = 1$ is trivial (both sides are $+\infty$). For $\sigma \neq 1$: let $\mathcal{O}_M = \mathcal{O}_L[\alpha]$ and $\mathcal{O}_L = \mathcal{O}_K[\beta]$. Then $e_{M/L} \cdot i_{L/K}(\sigma) = e_{M/L} \cdot v_L(\sigma(\beta) - \beta) = v_M(\sigma(\beta) - \beta)$, and $\sum_{\tau|_L = \sigma} i_{M/K}(\tau) = v_M\!\left(\prod_{g \in H}(\tau g(\alpha) - \alpha)\right)$ for any fixed $\tau$ with $\tau|_L = \sigma$. Set $b = \sigma(\beta) - \beta = \tau(\beta) - \beta$ and $a = \prod_{g \in H}(\tau g(\alpha) - \alpha)$. We show $v_M(a) = v_M(b)$ by proving $b \mid a$ and $a \mid b$ in $\mathcal{O}_M$. For $b \mid a$: let $F(x) \in \mathcal{O}_L[x]$ be the minimal polynomial of $\alpha$ over $L$, so $F(x) = \prod_{g \in H}(x - g(\alpha))$. Applying $\tau$ to the coefficients gives $(\tau F)(x) = \prod_{g \in H}(x - \tau g(\alpha))$. The difference $\tau F - F$ has coefficients of the form $\tau(z) - z$ for $z \in \mathcal{O}_L$, and any such $z$ satisfies $\tau(z) - z = \sigma(z) - z$ (as $\tau|_L = \sigma$), which is divisible by $b$ (since $b = \sigma(\beta) - \beta$ divides $\sigma(\beta)^i - \beta^i$ for all $i$, and $z \in \mathcal{O}_K[\beta]$). Evaluating at $x = \alpha$ gives $b \mid (\tau F - F)(\alpha) = \pm a$. For $a \mid b$: choose $f \in \mathcal{O}_K[x]$ with $f(\alpha) = \beta$. Then $f(x) - \beta = F(x)h(x)$ for some $h \in \mathcal{O}_L[x]$. Applying $\tau$: $(f - \tau\beta)(x) = (\tau F)(x)(\tau h)(x)$. Setting $x = \alpha$ gives $-b = \beta - \tau(\beta) = \pm a \cdot (\tau h)(\alpha)$, so $a \mid b$.