Why can we always restrict an Galois representation so that it becomes unramified? Let $K$ be a local field and $\rho: G_K \to \operatorname{GL}_n(\mathbb{C})$ be a Galois representation where $G_K$ denotes the absolute Galois group of $K$. 
We call a Galois representation $\rho$ unramified if $\rho(I_K)=1$ where $I_K$ denotes the inertia subgroup of $G_K$.

Question: Why does there exist a finite extension $K'/K$, so that the restriction of $\rho$ on $G_{K'}$ is unramified?

If $\rho$ would be an Artin representation, then, by definition, there exists a finite Galois extension $F/K$ such that $\operatorname{Gal}(\bar{K}/F) \subset \ker{\rho}$, i.e. $\rho$ comes from a representation $\operatorname{Gal}(F/K) \to \operatorname{GL}_n(\mathbb{C})$. Then $\rho|_F : \operatorname{Gal}(F/F) = \{1\} \to \operatorname{GL}_n(\mathbb{C})$ is obviously unramified.
But how can we approach the case of a general Weil representation? I heard that one can do this by noticing that $I_K$ is profinite and $W_K$ is the semidirect product of $I_K$ and $\mathbb{Z}$ where $\mathbb{Z}$ is the cyclic group generated by the Frobenius automorphism $x \mapsto x^{|k|}$ on $\bar{k}$, the algebraic closure of the residue field $k$ of $K$.
Could you please help me to explain this problem to me? Thank you in advance!
 A: $\DeclareMathOperator{\ur}{ur}$
$\DeclareMathOperator{\Gal}{Gal}$Let $K^{\ur}$ be the maximal unramified extension of $K$, so the inertia group $I_K$ is equal to $\Gal(\overline{K}/K^{\ur})$.  The restriction of $\rho$ to the compact group $I_K$ has open kernel, so there exists a finite Galois extension $E$ of $K^{\ur}$ such that $\rho|_{I_K}$ is well defined on $\Gal(E/K^{\ur})$.  
Claim 1: We can take $E$ to be the compositum of $K^{\ur}$ and a finite Galois extension $L$ of $K$.
Proof: If $E_1$ is another finite Galois extension of $K^{\ur}$ containing $E$, then we have $\Gal(\overline{K}/E_1) \subset \Gal(\overline{K}/E)$, so $\rho|_{I_K}$ is also well defined on $\Gal(\overline{K}/E_1)$.  Write $E = K^{\ur}(\beta)$, and take $L$ to be the Galois closure of $K(\beta)/K$.  Then $LK^{\ur}$ contains $E$ and does the trick.  $\blacksquare$
Claim 2: $L^{\ur}$, the maximal unramified extension of $L$ inside $\overline{L} = \overline{K}$, is equal to $E = LK^{\ur}$.
Proof: If $M$ is a finite unramified extension of $K$, then $LM$ is a finite unramified extension of $L$.  This shows $LK^{\ur} \subseteq L^{\ur}$.  The residue field of $LK^{\ur}$ is separably closed, giving us equality.  $\blacksquare$
Now consider the restriction of $\rho$ to the local Weil group $W_L$, which sits inside $\Gal(\overline{K}/L)$.  The local Weil group $W_L$ contains its inertia group $I_L = \Gal(\overline{K}/L^{\ur}) = \Gal(\overline{K}/LK^{\ur}) = \Gal(\overline{K}/E)$, on which $\rho$ is trivial.
