Number of elements in $D/P^e$ where $D$ is a ring of algebraic integers, and $P$ a prime ideal This is from Ireland and Rosen's A Classical Introduction to Modern Number Theory. 

Proposition 12.3.2: Consider a field $F/\mathbb Q$ with ring of integers $D$, and a prime ideal $P$ of $D$. Then the number of elements in $D/P^e$ is $p^{ef}$ where $D/P$ has $p^f$ elements.

The authors point out that $P^{e-1}/P^e$ is a subgroup of $D/P^e$ and has $p^f$ elements, and then go on to state that by induction the proposition is true. I do not see how induction works here.
 A: Here's basically the idea. Note that $D/P^e\cong D_P/P^eD_p$ so you might as well assume that $D$ is a DVR. Then, see this answer.
A: As Alex notes, you can replace $D$ by the localization $D_P$ without changing anything. Then $P$ becomes principal, and it is very easy to prove that $P^{r}/P^{r+1}$ is isomorphic to $D/P$ for all $r$. Now you can use the short exact sequences $$0\to P^{r}/P^{r+1}\to D/P^{r+1}\to D/P^{r}\to0$$ and induction to get what you want. Indeed.from this exact sequence we see that $$|D/P^{r+1}|=|P^{r}/P^{r+1}|\cdot|D/P^{r}|=|D/P|\cdot|D/P^{r}|$$
A: Since $D \supset P^{e-1} \supset P^e$, $|D/P^e| = |D/P^{e-1}| |P^{e-1}/P^e|$.
Hence, using induction on $e$,  it suffices to prove that $|P^{e-1}/P^e| = p^f$.
Since $P^e \subset P^{e-1}$ and $P^e \ne P^{e-1}$, there exist $\alpha \in P^{e-1} - P^e$.
We define a map $f\colon D \rightarrow P^{e-1}/P^e$ by $f(x) = \alpha x$ mod $P^e$.
This is a homomorphism of abelian groups.
Since $f(P) \subset P^e$, $f$ induces a homomorphism $g\colon D/P \rightarrow P^{e-1}/P^e$
It suffices to prove that $g$ is an isomorphism.
We first prove that $g$ is injective.
There exists an ideal $I$ such that $\alpha D = P^{e-1} I$.
$I$ is not divisible by $P$.
Suppose $f(x) \in P^e$, i.e. $\alpha x \in P^e$.
It suffices to prove that $x \in P$.
Suppose otherwise.
$\alpha xD = P^{e-1} I x$.
Since $I x$ is not divisible by $P$, $\alpha x$ is not divisible by $P^e$.
This is a contradiction.
Next we prove that $f$ is surjective.
It suffices to prove that $\alpha D + P^e = P^{e-1}$.
Since the only prime ideal containing $\alpha D + P^e$ is $P$,
$\alpha D + P^e$ is of the form $P^k$.
Clearly $k \ge e -1$.
Suppose $k \ge e$.
Then $\alpha \in P^e$.
This is a contradiction.
