Let $p$ be a prime number. Let $L/K/\mathbb{Q}_p$ be finite extensions where $L$ has relative ramification degree $e$ over $K$.
When I extend the $p$-adic absolute valute $|\cdot|$ to $L$ and $K$ I take the normalisation $|p| = 1/p$
Suppose we have $\mathcal{D}_{L/K} = \mathfrak{m}_L^l$ where $\mathcal{D}_{L/K}$ is the different ideal of $L/K$
I would like to show that $|Tr_{L/K}(x)| \leq p^{(e-l)/e)} |x|$ for all $x$ in $L$
I know that for any two fractional ideal $\mathfrak{a},\mathfrak{b}$ of $L$ and $K$ respectively we have the relation
$$ Tr(\mathfrak{a}) \subset \mathfrak{b} \Leftrightarrow \mathfrak{a} \subset \mathfrak{b} \mathcal{D}_{L/K}^{-1} $$
From which I conclude
$$ |Tr_{L/K}(x)| \leq p^{(e-l)/e)} |x| \Leftrightarrow |\mathcal{D}_{L/K}| \leq p^{(e-l)/e} $$
So if what I have done above is correct I left to show that
$$ |\pi_L|^l \leq p^{(e-l)/e)} $$
And for some reason I don't see why this is true. I feel it should be obvious but what I get is
$$|\pi_L|^l = p^{-l/ee'} $$ where $e'$ is the ramification index of $L$ over $\mathbb{Q}_p$. I'm obviously doing something wrong somewhere or missing something obvious so I would very much appreciate your help.