Clarification of proof that a graph $G$ of radius at most $k$ and maximum degree at most $d \ge 3$ has fewer than $\frac{d}{d - 2}(d - 1)^k$ vertices I am currently studying the textbook Graph Theory, fifth edition, by Reinhard Diestel. Chapter 1.3 Paths and cycles says the following:

Proposition 1.3.3. A graph $G$ of radius at most $k$ and maximum degree at most $d \ge 3$ has fewer than $\dfrac{d}{d - 2}(d - 1)^k$ vertices.
Proof. Let $z$ be a central vertex in $G$, and let $D_i$ denote the set of vertices of $G$ at distance $i$ from $z$. Then $V(G) = \bigcup_{i = 0}^k D_i$. Clearly $\lvert D_0 \rvert = 1$ and $\lvert D_1 \rvert \le d$. For $i \ge 1$ we have $\lvert D_{i + 1} \rvert \le (d - 1) \lvert D_i \rvert$, because every vertex in $D_{i + 1}$ is a neighbour of a vertex in $D_i$ (why?), and each vertex in $D_i$ has at most $d - 1$ neighbours in $D_{i + 1}$ (since it has another neighbour in $D_{i - 1}$). Thus $\lvert D_{i + 1} \rvert \le d(d - 1)^i$ for all $i < k$ by induction, giving
$$\lvert G \rvert \le 1 + d\sum_{i = 0}^{k - 1} (d - 1)^i = 1 + \dfrac{d}{d - 2}((d - 1)^k - 1) < \dfrac{d}{d - 2}(d - 1)^k. \square$$

I understood everything until this part:

and each vertex in $D_i$ has at most $d - 1$ neighbours in $D_{i + 1}$ (since it has another neighbour in $D_{i - 1}$). Thus $\lvert D_{i + 1} \rvert \le d(d - 1)^i$ for all $i < k$ by induction, giving
$$\lvert G \rvert \le 1 + d\sum_{i = 0}^{k - 1} (d - 1)^i = 1 + \dfrac{d}{d - 2}((d - 1)^k - 1) < \dfrac{d}{d - 2}(d - 1)^k. \square$$

Why does each vertex in $D_i$ have at most $d - 1$ neighbours in $D_{i + 1}$? How does it having another neighbour in $D_{i - 1}$ explain this?
How do we get $\lvert D_{i + 1} \rvert \le d(d - 1)^i$ for all $i < k$ by induction?
How does the author get
$$\lvert G \rvert \le 1 + d\sum_{i = 0}^{k - 1} (d - 1)^i = 1 + \dfrac{d}{d - 2}((d - 1)^k - 1) < \dfrac{d}{d - 2}(d - 1)^k$$?
 A: 
Why does each vertex in $D_i$ have at most $d - 1$ neighbours in $D_{i + 1}$? How does it having another neighbour in $D_{i - 1}$ explain this?

Each vertex in $D_i$ has degree at most $d$, because that is the maximum degree of $G$. So it has at most $d$ neighbors; if one of those neighbors is in $D_{i-1}$, at most $d-1$ of them can be in $D_{i+1}$.

How do we get $|D_{i + 1}| \le d(d - 1)^i$ for all $i < k$ by induction?

Our base case is $|D_1| \le d$, which is $d(d-1)^0$.
From there, assuming $|D_i| \le d(d-1)^{i-1}$ for some $i \ge 1$, we have $|D_{i+1}| \le (d-1)|D_i|$, so $|D_{i+1}| \le (d-1) \cdot d(d-1)^{i-1} = d(d-1)^i$.
By induction, this holds for all $i$.

How does the author get
$$\lvert G \rvert \le 1 + d\sum_{i = 0}^{k - 1} (d - 1)^i = 1 + \dfrac{d}{d - 2}((d - 1)^k - 1) < \dfrac{d}{d - 2}(d - 1)^k?$$

Assuming, as we do, that $G$ has radius at most $k$, every vertex of $G$ is contained in exactly one of the sets $D_0, D_1, D_2, \dots, D_k$. Therefore
$$|G| = |D_0| + |D_1| + \dots + |D_k| = |D_0| + \sum_{i=0}^{k-1} |D_{i+1}|.$$
Using $|D_0| = 1$ and $|D_{i+1}| \le d(d-1)^i$ for $i=0, \dots, k-1$, we get $$|G| \le 1 + \sum_{i=0}^{k-1} d(d-1)^i = 1 + d \sum_{i=0}^{k-1}(d-1)^i.$$
From here, we apply the formula $1 + r + r^2 + \dots + r^{k-1} = \frac{r^k - 1}{r-1}$ with $r = d-1$: the formula for the sum of a finite geometric series. This gives us
$$
  |G| \le 1 + d \cdot \frac{(d-1)^k - 1}{(d-1) - 1} = 1 + \frac{d}{d-2}((d-1)^k-1).
$$
The final step just cleans this up a little: expanding, then dropping a $-\frac{2}{d-2}$ term, we get
$$
   |G| \le 1 + \frac{d}{d-2} (d-1)^k  - \frac{d}{d-2} = \frac{d}{d-2}(d-1)^k - \frac{2}{d-2} < \frac{d}{d-2}(d-1)^k.
$$
