Let $p$ be a prime number. Let $n,m \geq 1$ be such that the topological spaces $\mathbb{Q}\_p^n$ 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 $n,m$.



p-adic noninvariance of dimension
$\mathbb{Q}\_p$ 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 $p$).


