Suppose I have a $\Gamma \simeq \mathbb Z_p $ extension $F_\infty /F$ of a number field $F$. Let $F_n$ be the fixed field of $\Gamma _n \simeq p^n \mathbb Z_p$. Denote by $L_n$ the maximal unramified abelian $p$ extension of $F_n$ in which all the primes above $p$ in $F_n$ split completely. Similarly, we may define $L_\infty$. Let $\mathcal L _n$ be the maximal abelian extension of $F_n$ contained in $L_\infty$.
We know that there exists an $n_0$ such that for $n\geq n_0$ all primes above $p$ are totally ramified in $F_\infty /F_n$. I want to prove that
$L_n \mathcal L_{n_0} = \mathcal L_n$.
It is easy to see that $L_n \mathcal L_{n_0}$ is an abelian extension of $F_n$. I don't know how to show that $L_n\mathcal L_{n_0} \subset L_\infty$ to conclude that $L_n \mathcal L_{n_0}\subset \mathcal L _n$. Also, how do I prove the other side?
Edit Is it true that $F_\infty =L_nF_\infty$?