ScalingStacks

Proof. In fact, let $\underline{s}=(s_{n})_{n\in\mathbb{N}}$ be an element of $V_{{\scriptscriptstyle\bullet}}(L)$ and $m\in\mathbb{N}$, let $n_{0}\in\mathbb{N}$ be the smallest integer for which $\vvvert\underline{s}\vvvert_{\phi}=\lVert s_{n_{0}}\rVert_{n_{0}\phi}$. By the ultrametricity of $\vvvert\mathord{\cdot}\vvvert_{\phi}$ and the power-multiplicativity of $\{\lVert\mathord{\cdot}\rVert_{n\phi}\}_{n\in\mathbb{N}}$, one has

$$\vvvert\underline{s}^{m}\vvvert_{\phi}\leqslant\lVert(s_{n_{0}})^{m}\rVert_{mn_{0}\phi}=(\lVert s_{n_{0}}\rVert_{n_{0}\phi})^{m}=\vvvert\underline{s}\vvvert_{\phi}^{m}.$$

By the choice of $n_{0}$, one has

$$\forall(j_{0},\dots,j_{l})\in\mathbb{N}^{l+1},\sum_{i\in\{0,\dots,l\}}i\cdot j_{i}=mn_{0},\quad\Big\lVert\prod_{i\in\{0,\dots,l\}}(s_{i})^{j_{i}}\Big\rVert_{mn_{0}\phi}\leq\lVert(s_{n0})^{m}\rVert_{mn_{0}\phi}$$

and the equality holds if and only if $(j_{0},\dots,j_{l})=(0,\dots,0,n_{0},0,\dots,0)$ where $n_{0}$ is on the $m$-th place, so by the definition of $\vvvert\mathord{\cdot}\vvvert$ and its ultra-metricity, one get

$$\vvvert\underline{s}^{m}\vvvert_{\phi}\geqslant\lVert(\underline{s}^{m})_{mn_{0}}\rVert_{mn_{0}\phi}\geq\lVert(s_{n0})^{m}\rVert_{mn_{0}\phi}=\vvvert\underline{s}\vvvert_{\phi}^{m},$$

hence there is an equality. ∎