In $\mathbb{Q}_p$ a closed ball of radius $p^n$ is the disjoint union of closed balls of radius $p^{n-1}$

Question

I try to understand why in $\mathbb{Q}_p$ a closed ball of radius $p^n$ is the disjoint union of closed balls of radius $p^{n-1}, n \in \mathbb{Z}$ or more precisely,

$\bar{B}(x, p^n) = \bigsqcup_{i=0}^{p-1} \bar{B}(x_i, p^{n-1})$ for some $x = x_0, x_1, \ldots, x_{p-1}$.

What I know so far is that open (closed) balls in $\mathbb{Q}_p$ are also closed (open) and that $\mathbb{Z}_p$ is the disjoint union of $p$ cosets of $p\mathbb{Z}_p$, i.e. of $p$ balls of radius $\frac{1}{p}$.

Can someone help me further ?

Thanks.