Note that $S$ is a closed point of $\tilde{\mathfrak{X}}^{\prime}$ and hence proper over $\tilde{K}$. Therefore also $Y$ is proper over $\tilde{K}$ since it is a closed subset of $\tilde{\iota}^{-1}(S)$ and $\iota$ is proper by [Tem00, Corollary 4.4]. Let $D$ be the Cartier divisor on $\mathfrak{X}^{\prime\prime}$ induced by $h$ as in Proposition 2.11 such that $c_{1}\left(\mathcal{O}(h\circ p_{\mathfrak{X}^{\prime}})\right)^{n}.Y=D^{n}.Y$. We show by induction that for all $0\leq l\leq n$ there is a strata cycle $Y_{l}$ of dimension $n-l$ whose components are contained in $Y$ such that $\Deg(D^{n}.Y)=\Deg(D^{n-l}.Y_{l})$. The case $l=0$ is clear by taking $Y_{0}:=Y$. Now let $l<n$ and $Y_{l}$ be as claimed. Let $Y^{\prime}$ be a stratum of $Y_{l}$, such that $Y^{\prime}$ is associated to an $l$-dimensional open face $\tau^{\prime}$ of $\mathfrak{D}$, i.e. $Y^{\prime}=\red_{\mathfrak{X}^{\prime\prime}}(p_{\mathfrak{X}^{\prime\prime}}^{-1}(\tau^{\prime}))$ with $u\in\overline{\tau^{\prime}}$ by the stratum face correspondence (Proposition 2.8). Using $\tau^{\prime}\subseteq\tau\subseteq\mathbb{R}^{n}$, there is an affine linear function $a:\mathbb{R}^{n}\rightarrow\mathbb{R}$ such that $h\Big|_{\tau^{\prime}}=a\Big|_{\tau^{\prime}}$. Then $h-a\Big|_{\tau}$ defines a Cartier divisor $D_{Y^{\prime}}$ on $\mathfrak{X}^{\prime\prime}$ by Proposition 2.11 which is numerically equivalent to $D$ on $Y$ by Lemma 2.13 and which is trivial on $Y^{\prime}$ because $h-a\Big|_{\tau^{\prime}}=0$. Hence, as $\overline{Y^{\prime}}$ is a strata subset, $D_{Y^{\prime}}.\overline{Y^{\prime}}$ is a strata cycle. Write $Y_{l}=\sum_{Y^{\prime}}m_{Y^{\prime}}\overline{Y^{\prime}}$ where the sum ranges over a finite number of $n-l$-dimensional strata of $\tilde{\mathfrak{X}}^{\prime\prime}$ contained in $Y$. Then we can calculate:

	$\displaystyle\Deg(D^{n}.Y)$	$\displaystyle=\Deg\left(D^{n-l}.Y_{l}\right)$
		$\displaystyle=\Deg\left(D^{n-l}.\sum_{Y^{\prime}}m_{Y^{\prime}}\overline{Y^{\prime}}\right)$
		$\displaystyle=\Deg\left(\sum_{Y^{\prime}}m_{Y^{\prime}}D^{n-l}.\overline{Y^{\prime}}\right)$
		$\displaystyle=\Deg\left(\sum_{Y^{\prime}}m_{Y^{\prime}}D^{n-l-1}.(D_{Y^{\prime}}.\overline{Y^{\prime}})\right)$
		$\displaystyle=\Deg\left(D^{n-l-1}.\sum_{Y^{\prime}}m_{Y^{\prime}}D_{Y^{\prime}}.\overline{Y^{\prime}}\right)$

and $Y_{l+1}:=\sum_{Y^{\prime}}m_{Y^{\prime}}D_{Y^{\prime}}.\overline{Y^{\prime}}$ is a strata cycle as claimed. We use this for $l=n$ to see that $\Deg(D^{n}.Y)=\Deg(Y_{n})$ for a strata cycle $Y_{n}$ of dimension $0$ contained in $Y$. Its components are strata points $S_{i}$ of $\mathfrak{X}^{\prime\prime}$ which are mapped by $\iota$ to the point $S$ corresponding to $\tau$. Now let $\mathfrak{U}^{\prime}\subseteq\mathfrak{X}^{\prime}$ be a formal open subset with an étale morphism $\psi:\mathfrak{U}^{\prime}\rightarrow\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})$ such that $S$ is the distinguished stratum of $\mathfrak{U}^{\prime}$ (cf. Proposition 2.5) and define $\mathfrak{U}^{\prime\prime}:=\iota^{-1}(\mathfrak{U}^{\prime})$. Note that there is no factor $\mathfrak{X}(m)$ because $\tau$ is of maximal dimension. As the strata occurring in the intersection process correspond to open faces of $\mathfrak{D}$ with vertex $u$, their intersection with $\mathfrak{U}^{\prime\prime}$ is nonempty. Hence we may calculate the multiplicities of $Y_{n}$ locally on $\mathfrak{U}^{\prime\prime}$. The stratification of $\tilde{\mathfrak{U}}^{\prime\prime}$ is obtained by the preimages of the strata of $\tilde{\mathfrak{X}}(\boldsymbol{n},\boldsymbol{a})^{\prime}$ (see proof of Proposition 2.8) with respect to the base change $\psi^{\prime}:\mathfrak{U}^{\prime\prime}\rightarrow\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})^{\prime}$ of $\psi$ (cf. Construction 2.6). Let $Y_{u}=\overline{\tilde{\psi}^{\prime}(\tilde{\mathfrak{U}}^{\prime\prime}\cap Y)}$ be the irreducible component in $\tilde{\mathfrak{X}}(\boldsymbol{n},\boldsymbol{a})^{\prime}$ corresponding to $u$ and $D_{u}$ the Cartier divisor on $\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})^{\prime}$ whose pullback gives the Cartier divisor $D$ associated to $h$ on $\mathfrak{U}^{\prime\prime}$ (cf. proof of Proposition 2.11). By applying the modifications of $D$ in the induction step also to $D_{u}$ we obtain a strata cycle $Y_{n}^{t}=\sum m_{j}P_{j}$ of $\tilde{\mathfrak{X}}(\boldsymbol{n},\boldsymbol{a})^{\prime}$ whose pullback is $Y_{n}$ (as the intersection product is compatible with flat pullback by [Ful98, Proposition 2.3(d)]) and which has the same degree as $D_{u}^{n}.Y_{u}$ using Lemma 2.13. Now let

	$\displaystyle\val:(\mathbb{G}_{m}^{\boldsymbol{n}})_{K}^{\textup{an}}$	$\displaystyle\rightarrow\mathbb{R}^{\boldsymbol{n}},$
	$\displaystyle q$	$\displaystyle\mapsto(-\log q(x_{01}),...,-\log q(x_{0n_{0}}),...,-\log q(x_{p1}),...,-\log q(x_{pn_{p}}))$

and $\Sigma:=\left\{\boldsymbol{w}\in\mathbb{R}_{\geq 0}^{\boldsymbol{n}}\;\Big|\;w_{i1}+...+w_{in_{i}}\leq v(a_{i}),0\leq i\leq p\right\}$. As we have an isomorphism

$$\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})^{\textup{an}}\tilde{\rightarrow}\val^{-1}(\Sigma)$$

and using [Gub13, Corollary 6.15], we find that $Y_{u}$ is a toric variety with fan given by the cones generated by $\Delta^{\prime}-u$ for $\Delta^{\prime}\in\mathfrak{D}$ with vertex $u$ (in fact we identify $\tau$ with $\Sigma$ by forgetting about the coordinate with index $0$ for each $i$). $D_{u}\Big|_{Y_{u}}$ is given up to multiplication by a constant by the divisor $D^{\prime}_{u}$ on $Y_{u}$ associated to the linear function $h^{\prime}:=h(\cdot+u)-h(u)$. By [Ful93, 3.4,5.3] we have

$$\lambda(P_{D^{\prime}_{u}})=\frac{\Deg({D^{\prime}}_{u}^{n}.Y_{u})}{n!},$$

where

$$P_{D^{\prime}_{u}}=\left\{y\in\mathbb{R}^{n}\;\Big|\;\langle z,y\rangle\leq\psi_{D^{\prime}_{u}}(z)=h^{\prime}(z)\;\forall z\in\mathbb{R}^{n}\right\}=\nabla h^{\prime}(0)$$

and $\lambda$ denotes the standard Lebesgue measure. For the last term we get

	$\displaystyle\nabla h^{\prime}(0)$	$\displaystyle=\left\{p\in\mathbb{R}^{n}\;\Big|\;\forall x\in\tau-u\;:\;h^{\prime}(0)+\langle x,p\rangle\leq h^{\prime}(x)\right\}$
		$\displaystyle=\left\{p\in\mathbb{R}^{n}\;\Big|\;\forall x\in\tau-u\;:\;\langle x,p\rangle\leq h(x+u)-h(u)\right\}$
		$\displaystyle=\left\{p\in\mathbb{R}^{n}\;\Big|\;\forall x\in\tau\;:\;h(u)+\langle x-u,p\rangle\leq h(x)\right\}$
		$\displaystyle=\nabla h(u).$

Hence

$$\frac{1}{n!}\Deg(Y_{n}^{t})=\frac{\Deg({D^{\prime}}_{u}^{n}.Y_{u})}{n!}=\lambda(P_{D^{\prime}_{u}})=\lambda(\nabla h(u))=\MA(h)(\{u\}).$$

With $\iota^{\prime}$ denoting the morphism $\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})^{\prime}\rightarrow\mathfrak{X}(\boldsymbol{n},\boldsymbol{a})$ we conclude

$$\Deg(D^{n}.Y)=\Deg\left(Y_{n}\right)=\Deg\left(\psi^{\prime\ast}Y_{n}^{t}\right)=\Deg\left(\iota_{\ast}\psi^{\prime\ast}\sum m_{j}P_{j}\right).$$

Using [Ful98, Proposition 1.7] this equals

$$\Deg\left(\psi^{\ast}\iota^{\prime}_{\ast}\sum m_{j}P_{j}\right)=\Deg\left(\psi^{\ast}\sum m_{j}[P_{j}:\{\tilde{\boldsymbol{0}}\}]\cdot\{\tilde{\boldsymbol{0}}\}\right).$$

As $\psi^{-1}(\{\tilde{\boldsymbol{0}}\})=S$ is reduced since $\psi$ is smooth, this amounts to

$$\Deg\left(\sum m_{j}\Deg(P_{j})S\right)=\Deg(S)\cdot\Deg(Y_{n}^{t})=\Deg(S)\cdot n!\cdot\MA(h)(\{u\}).$$

This yields the equality we wanted to prove. ∎