An speical generating element of the ring of integers in local fields See the accepted answer to this question Can a generator of the ring of integers of local fields can be chosen so that it is also a uniformizer at the same time?
"$O_L=O_K[\pi_L]$ iff $f(L/K)=1$ and $O_L=O_K[\zeta_{q_L-1}]$ iff $e(L/K)=1$." I can see why this sentence is true.
"In general it is $O_L=O_K[\zeta_{q_L-1}+\pi_L]$." I can not see it. I can just see that $O_L=O_K[\zeta_{q_L-1}, \pi_L]$, and $O_K[\zeta_{q_L-1}+ \pi_L] \subseteq O_K[\zeta_{q_L-1}, \pi_L]$
 A: I have a proof but this may not be the same one as suggested by reuns in the link.
Let me try to restate the claim:

Let $l,k$ be residue fields of finite extension of local fields $L/K$ and let $\mathfrak{q},\mathfrak{p}$ be corresponding maximal ideals of $\mathcal{O}_L,\mathcal{O}_K$. Then $l/k$ is extension of finite fields, hence $l=k(\zeta_{q^n-1})$ where $q=|k|$ is a prime power, $q^n=|l|$ and $\overline{g}=x^{q^n-1}-1$ minimal polyonomial of $\zeta_{q^n-1}$.  We have $\mathcal{O}_L=\mathcal{O}_K[\zeta_{q_L-1}+\pi_L]$ where here we interpret $\zeta_{q_L-1}$ as any lift of $\zeta_{q^n-1}\in l$ to $\mathcal{O}_L$.

Let $a_0$ be any lift of $\zeta_{q^n-1}\in l$ to $\mathcal{O}_L$ with minimal polynomial $g\in \mathcal{O}_K[x]$ whose taking mod $\mathfrak{p}$ is $\overline{g}$. We let $a=\begin{cases}a_0 & v_{\mathfrak{q}}(g(a_0))=1 \\ a_0+\pi_L & v_{\mathfrak{q}}(g(a_0))>1 \end{cases}$ and we will show that $\mathcal{O}_L=\mathcal{O}_K[a]$.
First, the choice of $a$ gives $v_{\mathfrak{q}}(g(a))=1$. Indeed, consider when $v_{\mathfrak{q}}(g(a_0))>1$, $a=a_0+\pi_L$. We can expand $g(x)=g(a_0)+(x-a_0)g'(x)+h(x)(x-a_0)^2$ for $h\in \mathcal{O}_K[x]$ (here $g'$ is the formal derivative of $g$) then $g(a)=g(a_0)+\pi_L g'(a)+h(a)\pi_L^2$. Hence, we just need to show $v_{\mathfrak{q}}(g'(a))=0$. Assume otherwise, then $0\equiv g'(a)\equiv g'(a_0)\pmod{\mathfrak{q}}$ so this implies $\overline{g}$ and its derivative $\overline{g}'$ has $\zeta_{q^n-1}$ as common root, a contradiction since $\overline{g}=x^{q^n-1}-1$ is separable.
Now, with $[L:K]=m=ne$ then we want to show $1,a,\ldots, a^{m-1}$ generate $\mathcal{O}_L$ as $\mathcal{O}_K$-module. By Nakayama's lemma (Version 3), it suffices to show $1,a,\ldots, a^{m-1}$ spans $\mathcal{O}_L/\mathfrak{p}$ as $k=\mathcal{O}_K/\mathfrak{p}$-vector space. As $\mathfrak{p}\mathcal{O}_L=\mathfrak{q}^e$ so every element of $\mathcal{O}_L/\mathfrak{p}$ can be written as $b_0+b_1\pi_L+b_2\pi_l^2+\ldots + b_{e-1}\pi_L^{e-1}$ where $b_i\in l=\mathcal{O}_L/\mathfrak{q}$ and $\pi_L$ uniformiser of $\mathcal{O}_L$, which we can choose to be $g(a)$ because $v_{\mathfrak{q}}(g(a))=1$. Because $\pi_L=g(a)$ and $l=k(\zeta_{q^n-1})$ where $a\pmod{\mathfrak{q}}=\zeta_{q^n-1}$. Hence, each $b_i$ can be written as linear combination of $a\pmod{\mathfrak{q}}$ and $\pi=g(a)$, a polynomial in $a$. We are done.
Observe that when $e=1$ then we don't need the condition $v_{\mathfrak{q}}(g(a))=1$ to show $\mathcal{O}_L=\mathcal{O}_K[a]$, hence in this case, $a$ can be any lift of $\zeta_{q^n-1}$.
