If $K$ is a finite extension of $\mathbb{Q}_p$ degree $n$, index of ramification $e$, and residue field degree $f$, and if $\pi$ chosen so that $\text{ord}_p \pi=\frac{1}{e}$, then every $\alpha \in K$ can be written in one and only one way as $$ \alpha=\sum_{i=m}^{\infty}\alpha_i \pi^i,$$ where $m=e \cdot \text{ord}_p \alpha$ and $\alpha_i$ satisfies $a_i^{p^{f}}=\alpha_i$ i.e., $\alpha_i$' s are Teichm$\ddot{u}$ller digits.
Answer:
I have used this result but I have no proof available to me.
I have thought like the following:
Since $ \text{ord}_p(\pi)=\frac{1}{e}$, for any $\alpha \in K$, we have
$ \alpha=\pi^m u, ..............(1)$
where $u$ is a unit in $K$ and $m=e \cdot \text{ord}_p(x)$
Also we have $n=ef$.
Using Hensel's lemma we can derive
$\alpha \equiv \alpha_0 \ (mod \ \pi), \\ \alpha \equiv \alpha_0+\alpha_1 \pi \ (mod \ \pi^2), \\ \alpha \equiv \alpha_0+\alpha_1 \pi+\alpha_2 \pi^2 \ (mod \ \pi^3), \\ ..... \\ \alpha=\alpha_0+ \alpha_1 \pi+\alpha_2 \pi^2+..........$, where $ |\alpha-\alpha_0|_p<1$
This shows that $ \alpha=\sum_{i=0}^{\infty} \alpha_i \pi^i, ............(2)$
I think from $(1)$ and $(2)$, we can prove but I can not prove it.
I need your help.