Question

let $p$ be a prime. let $n,m \geq 1$ such that $\mathbb{Q}_p^n$ and $\mathbb{Q}_p^m$ are homeomorphic. can we conclude $n=m$?

for $\mathbb{Z}_p$ it's false: in fact, brouwer's theorem implies that $\mathbb{Z}_p$ is homeomorphic to the cantor set $C$, which of course satisfies $C^n \cong C^m$ for all $n,m$.

Answer 1

$\mathbb{Q}_p$ is homeomorphic to a countable direct sum of copies of the Cantor set $C$. Indeed, because the valuation is discrete, for each $n \geq 1$ the "annulus"

$A_n =$ {$x \in \mathbb{Q}_p \ | \ p^{n-1} < ||x|| \leq p^{n}$}

is closed and homeomorphic to the Cantor set $C$. (Take of course $A_0 = \mathbb{Z}_p$.)

Since as you observed above, $C \times C \cong C$, it follows that $\mathbb{Q}_p^n$ and $\mathbb{Q}_p^m$ are homeomorphic for all $m, \ n \in \mathbb{Z}^+$ (and the homeomorphism type is independent of $p$).