# What is the cardinality of $\mathcal{O}_L/\pi_L^n \mathcal{O}_L$ for some $n \in \mathbb{N}$?

Let $$K \supset \mathbb{Q}_p$$ be the $$p$$-adic field with ring of integers $$\mathcal{O}_K$$ and $$\pi_K$$ be its uniformizer . Let $$L$$ be an unramified extension of $$K$$ of dgree $$d$$ and ring of integers $$\mathcal{O}_L$$. Let $$\pi_L$$ be an uniformizer of $$\mathcal{O}_L$$.

What is the cardinality of $$\mathcal{O}_L/\pi_L^n \mathcal{O}_L$$ for some $$n \in \mathbb{N}$$ ?

Since $$L$$ is an unramified extension of degree $$d$$, we have $$[ \mathcal{O}_L/\pi_L \mathcal{O}_L:\mathcal{O}_K/\pi_K \mathcal{O}_K]=d$$, where $$q=| \mathcal{O}_K/\pi_K \mathcal{O}_K|$$.

Therefore $$|\mathcal{O}_L/\pi_L^n \mathcal{O}_L|=(q^d)^n=q^{dn}$$.

Am I correct ?

Namely, if we abbreviate the residue fields as $$l := \mathcal{O}_L/\pi_L \mathcal{O}_L$$ and $$k := \mathcal{O}_K/\pi_K \mathcal{O}_K$$, then the extension being unramified of degree $$d$$ means more or less by definition that $$[l:k]=\color{red}{d}$$ (not $$q^d$$). Together with $$q := card(k)$$ this of course gives $$card(l) = q^d$$.
So that would solve the case $$n=1$$. To conclude for $$n \ge 2$$, I suggest induction, and using certain exact sequences. But maybe some other method would work too. Can you fill in these details?