Let $k$ be a $p$-adic field with residue field $\kappa$ of cardinality $q$ and characteristic $p$.
The Weil group $W(k)$ is the subgroup of $\operatorname{Gal}(\overline{k}/k)$ of automorphisms that reduce to an integral power of Frobenius $x\mapsto x^q$ in $\operatorname{Gal}(\overline{\kappa}/\kappa)$. We have the exact sequence $1\to I\to W(k)\to \mathbf{Z}\to 1$, where $I$ is inertia and $\mathbf{Z}$ is the subgroup generated by the image of a Frobenius element $F$ in $\operatorname{Gal}(\overline{k}/k)$.
Why is this short exact sequence split, i.e. why is $W(k)=I\rtimes \langle F\rangle$, and what is the action of $F$ on $I$? Is it raising to the power $q$? I read that the conjugation action is given by raising to the power $q$ modulo the wild inertia group $I_+$.
The maximal tame extension $k_t$ is generated by adjoining to the maximal unramified extension $k_u$ all $\sqrt[e]{\pi}$ with $(p,e)=1$ and $\pi$ a uniformiser of $k$. With this description, it is easy to see that $\operatorname{Gal}(k_t/k_u)=\prod_{p\neq l}\mathbf{Z}_\ell$. Somewhere else, I read that $k_t=k_u(\sqrt[q^n-1]{\pi}:n\geq 1)$. With this description, it is easy to see that $\operatorname{Gal}(k_t/k_u)=\varprojlim_n \mathbf{F}_{q^n}^\times$. Why are these two equivalent? Not every $e$ coprime with $p$ is of the form $q^n-1$, right?
Any help is much appreciated.