Proof help- the set $\{ P \in \mathbb{P}^N(\bar{\mathbb{Q}}) : H(P) \leq C, [\mathbb{Q}(P):\mathbb{Q}] \leq d] \}$ is finite ((Heights in Silverman) I have a doubt in the following theorem (Silverman AEC)

Doubt 1: Why is the expression for $$H_{\mathbb{Q}(P)}(P) \geq max_{ 0 \leq i \leq N} \Big(\prod_{v \in \mathbb{Q}(P)} max\{|x_i|_v,1\}^{n_v} \Big)?$$
I tried it for $N = 0$ case (to see why it could be true): then LHS simplifies to $\prod_{v \in \mathbb{Q}(P)} |x_0|_v^{n_v}$ while RHS (the expression on the right of $\geq $) equals $\prod_{v \in \mathbb{Q}(P)} max\{|x_0|_v,1\}^{n_v}$. 
We can split this product into those $v \in \mathbb{Q}(P)$ for which $|x_0|_v \geq 1$ and $|x_0| \leq 1$. But by the choice of homogeneous coordinates that at least one $x_j=1$, we have LHS = RHS here. 
And how do I generalise this reasoning? (I guess that wouldn't be hard once I have at least $N =0,1$ case down) 
Doubt 2: The book says it is sufficient to prove consider the case $N =1$. How do I show that? 
For any $N$, and fixed $C, d$, by $N=1$ case we have $H_{\mathbb{Q}(x_i)}(x_i) \leq C$ and $[\mathbb{Q}(x_i):\mathbb{Q}] \leq d$ for all $ i \in \{ 0, 1, ... , N\}$. 
Then $max_{ 0 \leq i \leq N} H_{\mathbb{Q}(x_i)}(x_i) \leq C^d$ and $max_{0 \leq i \leq N} [\mathbb{Q}(x_i): \mathbb{Q}] \leq d$. 
Am I on the right track? Would really appreciate any help with this. 
Thank you very much. 
 A: Doubt 1: The inequality he wants to show is
$$H(P) = \prod_v \max_i\{|x_i|_v\}^{n_v} \geq \max_i\left\{\prod_v \max\{|x_i|_v,1\}^{n_v}\right\}$$
On the right hand side, let's say the maximum is achieved by $i=k$ fixed. So now the RHS is just
$$\prod_v \max\{|x_k|_v,1\}^{n_v}$$
And this we can compare termwise to the LHS. So we reduce to showing that
$$\max_i \{|x_i|_v\} \geq \max\{x_k,1\}^{n_v}$$
Recall that we normalized $P$ so that one of the $x_i$'s is $1$, and $|1|_v = 1$ for all $v$. So on the left, the set we take the maximum over includes both $1$ and $x_k$, but also the other $x_i$'s so it can only be bigger than the maximum of $1$ and $x_k$. 
p.s. the $N=0$ case won't help much because $\mathbb P^0$ doesn't really exist, and if it did it would probably have just 1 point.
Doubt 2: as it says in the book, the reduction to $N=1$ is essentially done by applying that case to each (dehomogenized) projective coordinate.
If we show that there are only finitely many choices of each $x_i$ in this setting, then we are done, because then there are only finitely many choices for $P$. 
The $N=1$ case can be applied to the points $[x_i:1]$ in $\mathbb P$ if we show that $(1)$ $H([x_i:1])$ is bounded and $(2)$ that the degree of $\mathbb Q([x_i:1])$ is bounded.
(1) By the first inequality, 
$$H(P) \geq \max H(x_i)$$
so each $H([x_i:1])$ is bounded by the same bound on $H(P)$.
(2) Observe that after normalizing $P$ to make one of its coordinates $1$, we have essentially dehomogenized it, and in particular $\mathbb Q([x_i:1]) = \mathbb Q(x_i) \subseteq \mathbb Q(P)$, and so if the degree of the latter is bounded, so is the degree of the former.
Thus for each coordinate $x_i$, there are only finitely many choices, and hence finitely many points $P$.
I don't quite follow your work for this case. I can see that you have extra exponents associated to the degrees of $\mathbb Q(x_i)$, but remember that the global height is defined so that thsoe don't appear. The inequality also seems like it is in the wrong direction, or something like that.
