TITLE: Degree of closure of $\mathbb{Q}_p$
QUESTION [7 upvotes]: In order to prove that algebraic closure of $\mathbb{Q}_p$ is infinite, I took the polynomial $x^n-p$ with $n>1$ over $\mathbb{Q}_p$ to show that this eqaution has no solution for infinite cases to get the result but I want to make one part of the proof precise. 
It's clear that $x^2-p$ has no solution in the field so we adjoin $\sqrt{p}$ to get $\mathbb{Q}_p(\sqrt{p})$. For higher powers of $n$, clearly $p^{1/n}$ is not contained in $\mathbb{Q}_p$ but how do we show that $p^{1/n}$ is not contained in the extensions with smaller $n$? For instance how do we show that $\sqrt[3]{p}\notin \mathbb{Q}_p(\sqrt{p})$?

Addendum:
Reading Prof. Conrad's comment is helpful for the discussion.

REPLY [5 votes]: I assume you mean that you want to show the algebraic closure is of infinite degree over $\mathbf{Q}_p$. It's then enough to show that $\mathbf{Q}_p$ admits finite extensions of arbitrary degree (if an algebraic closure of $\mathbf{Q}_p$ has finite degree $n$, note that it contains a finite extension of degree larger than $n$, a contradiction). For this it's enough to observe that $x^n-p$ is irreducible over $\mathbf{Q}_p$ for all $n\geq 1$ by Eisenstein's criterion ($p$ is prime in the UFD $\mathbf{Z}_p$).