The algebraic closure of $\mathbb{Q}_p$ I am trying to explain why $\mathbb{Q}_p^{\text{alg cl}}$ is an infinite field extension of $\mathbb{Q}_p$ (unlike $\mathbb{C}/\mathbb{R}$ which has deg 2).
Does the following argument work out...
If $p=2,$ then for all $n \in \mathbb{N}$ the polynomial $X^{2n}-2$ has no roots (mod $4$) and so has none in $\mathbb{Q}_2.$ 
Moreover, $X^{2n}-2$ is Eisenstein at $2$ and so it is irreducible over $\mathbb{Q}_2.$
Hence $\mathbb{Q}_2^{\text{alg cl}}$ is an infinite field extension of $\mathbb{Q}_2.$
As for $p\neq 2,$ we choose a quadratic non-residue (mod $p$) and proceed as above. Q.E.D.
Many thanks!
 A: Your argument for the case $p=2$ works, but it can be simplified: By Eisenstein, any polynomial $$X^n+2, \, n>2$$ is irreducible and gives rise to a field extension of degree $n$. 
In the case $p\neq 2$ we have to be careful: Just choosing quadratic non-residues is not enough for higher powers, since for example $3$ is a quadratic non-residue $\bmod 7$, but $$X^4-3=(X^2+2X+2)(X^2+5X+2) \bmod 7$$ isn't irreducible. However, you can use just the same idea as above with $2$ replaced by $p$.
By the way: The Artin-Schreier-Theorem states that if $C$ is an algebraic closure of a field $K$ with $C|K$ finite, then $C$ must be the splitting field of $X^2+1$ over $K$ and $K$ is of characteristic $0$.
A: Your proof for $p=2$ certainly works, though the key is Eisenstein. You don't need to show that $X^{2n}-2$ has no roots (that wouldn't be enough anyway to prove what you want).
In general for arbitrary $p$, note that $X^n-p$ is Eisenstein in $\mathbb{Q}_p$, and so its irreducible. By varying $n$ you find that $\mathbb{Q}_p$ has algebraic field extensions of arbitrarily high degree.
