Completion of the maximal unramified extension of the field of $p$-adic numbers $\newcommand{\Q}{\mathbb{Q}}
\DeclareMathOperator{\unr}{unr}
\newcommand{\C}{\mathbb{C}}
$
Denote by $\Q_p^{\unr}$ the maximal unramified extension of the field of $p$-adic numbers $\Q_p$ (where $p$ is any prime). The $p$-adic absolute value on $\Q_p$ has a unique extension to $\Q_p^{\unr}$, which I denote $| \cdot |_p$.
I would like to know what is the completion $K$ of $\Q_p^{\unr}$ with respect to $| \cdot |_p$. It is known that $K \neq \Q_p^{\unr}$. I wonder if $K = \C_p$ holds, where $\C_p$ denotes the completion of the algebraic closure $\overline{\Q_p}$ of $\Q_p$, or equivalently the completion of $\overline{\Q}$ with respect to $| \cdot |_p$. So here is my precise question :

What is the degree $[\C_p : K]$ equal to ?

By Artin–Schreier's theorem and because $\C_p$ is algebraically closed, if $1 < [\C_p : K] < \infty$, then $K$ would be a real closed field, which is not the case because $K$ contains non-trivial roots of unity. Therefore, the degree $[\C_p : K]$ is either $1$ or is some infinite cardinal.
Any idea about my question will be appreciated.
 A: $\newcommand{\Q}{\Bbb Q}
\newcommand{\trdeg}{\mathrm{tr.deg}}
\newcommand{\C}{\Bbb C}
\newcommand{\unr}{\rm unr}$
The answer is given as theorem 3 in the paper David Lampert, Algebraic $p$-adic expansions, Journal of Number Theory, volume 23 (3), 1986, pp. 279-284.
Consider the tower of extensions, where $K = \widehat{ \Q_p^{\unr} }$ :
$\hspace{6cm}$
We have, according to the cited paper :
$$\trdeg_{\Q_p}(K) = \trdeg_K(\C_p) = 2^{\aleph_0}.$$
Since $\#\C_p = \#\Q_p = \# K = 2^{\aleph_0},$ it is easy to conclude that $[\C_p : K] = 2^{\aleph_0}$ (this is the algebraic degree – no Hilbert basis or whatever).
A: Let me try to answer to the question - as modified by @Lubin - not in terms of "infinite degrees" (perhaps too vague a notion) but in terms of cardinals of Galois groups. As a base field, take an ultrametric complete field $K$, say a finite extension of $\mathbf Q_p$ for simplification, and an algebraic closure $\bar K$ with Galois group $G_K=Gal(\bar K/K)$ and completion $\hat {\bar K}$. It is known that any automorphism in $G_K$ acts as an isometry hence its action can be extended to $\hat {\bar K}$ by continuity. For a closed subgroup $H$ of $G_K$, let $L$ be the fixed field of $H$, so that $H=G_L=Gal(\bar K /L)$ . In the OP question, $K=\mathbf Q_p$  and $L=\mathbf Q_p^{nr}$. 
A central result in this context is the Ax-Sen-Tate theorem (*), which states states that $\hat {L}=(\hat {\bar K})^H$, so in particular $H=Gal(\hat {\bar K}/\hat {L})$. Take $L= K^{nr}$, so $H=G_{K^{nr}}$. Replacing the "degree" of  $\hat {\bar K}/\hat {L}$ by card $H$, we want to show that $H$ is not enumerable. The advantage of the AST theorem is to bring us back to "usual" algebraic extensions (not necessarily complete). The algebraic closure $\bar K$ contains the maximal abelian extension $K^{ab}$ of $K$, and so $G_{K^{nr}}$ surjects onto $Gal(K^{ab}/K^{nr})$, which is known by local CFT to surject onto $\mathbf Z_p$, see e.g. Cassels-Fröhlich, chap.VI, §2.8 (in the case $K=\mathbf Q_p$, this is just the local Kronecher-Weber theorem, op. cit. §3.1). But card $\mathbf Z_p=\aleph_1$, and we are done .
(*) J. Tate, "$p$-divisible groups", Proc. Conf. Local Fields, 1966
