Degree of closure of $\mathbb{Q}_p$ 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. 

The $p$-adic valuation extends uniquely to any finite extension of $\mathbf Q_p$.
Remark that $v_p(p^{1/n})=1/n$, so $v_p(\mathbf Q_p(p^{1/n})^\times) = 1/n\cdot\mathbf Z$. In particular, the field $\mathbf Q_p(p^{1/n})$ cannot be embedded in $\mathbf Q_p(p^{1/m})$ unless $n|m$.

If you just want to show that $\overline{\mathbf Q_p}$ is not finite over $\mathbf Q_p$, then Keenan's method is the easiest. But there's a funny way to hit this simple statement very hard with the theory of real closed fields. According to a theorem, if $F$ is a non-algebraically closed field whose algebraic closure $\overline{F}$ is finite over $F$, then $F$ is elementarily equivalent to the real numbers. Since $x^2-p$ splits over $\mathbf R$ but not over $\mathbf Q_p$, it follows that $\overline{\mathbf Q_p}/\mathbf Q_p$ is an infinite extension!
