Convergence of $\sum_{\alpha\in \Bbb N^n} z^\alpha$ in $\{z\in \Bbb C^n: |z_i| < 1 \text{ for all } 1\le i\le n\}$ I am reading Introduction to Complex Analysis in Several Variables by Volker Scheidemann.
The convergence of power series in several complex variables is defined in the following way:

Let $\{c_\alpha\}_{\alpha\in\Bbb N^n} \subset\Bbb C$. The power series $\sum_{\alpha\in \Bbb N^n}c_\alpha (z-a)^\alpha = \sum_{\alpha\in \Bbb N^n} c_\alpha (z_1-a_1)^{\alpha_1} \ldots (z_n-a_n)^{\alpha_n}$ in $n$ variables $z:= (z_1,z_2,\ldots,z_n)$ centered at $a\in \Bbb C^n$ converges at $w\in \Bbb C^n$ if there exists $C> 0$ such that $$\sum_{\alpha\in F} |c_\alpha| |(w-a)^\alpha| \le C$$ for all finite subsets $F\subset\Bbb N^n$.

I am trying to understand the convergence behavior of $\sum_{\alpha\in \Bbb N^n} z^\alpha$ in the unit polycylinder $P^n_1(\mathbf 0) := \{z\in \Bbb C^n: |z_i| < 1 \text{ for all } 1\le i\le n\}$. This is Example $1.5.7$ of the book.

Let $z\in P^n_1(0)$ and $F\subset \Bbb N^n$ be finite. There exists $q\in [0,1)$ such that $|z_j| \le q$ for all $1\le j\le n$. Hence,
$$\sum_{\alpha\in F} |z^\alpha| = \sum_{\alpha\in F} |z_1|^{\alpha_1} \ldots |z_n|^{\alpha_n} \le \color{blue}{\sum_{\alpha\in F} q^{\alpha_1 + \ldots + \alpha_n} \le \sum_{j=0}^\infty q^j} = \frac{1}{1-q}$$

Why is the inequality in blue true? Certainly, there may exist distinct $\beta,\gamma\in F$ such that $\sum_{i=1}^n \beta_i = \sum_{i=1}^n \gamma_i$. In such a case, I don't see how to get the above inequality.
Perhaps the homogeneous expansion may be useful at some point: $$\sum_{\alpha\in \Bbb N^n} c_\alpha (w-a)^\alpha = \sum_{j=0}^\infty \left(\sum_{|\alpha| = j} c_\alpha (w-a)^\alpha \right)$$
In fact, using the above homogeneous expansion (once convergence issues are settled) the author goes on to say that

$$\color{blue}{\sum_{k=0}^\infty \left(\sum_{|\alpha| = k} z^{\alpha} \right) = \sum_{k_1 = 0}^\infty \ldots \sum_{k_n = 0}^\infty z_1^{\alpha_1} \ldots z_n^{\alpha_n}}$$

Why is this true?
Notation. For $\alpha\in \Bbb N^n$, $|\alpha|$ denotes $\alpha_1 + \ldots + \alpha_n$.
 A: You're correct - it is not true that
$$ \sum_{\alpha \in F} q^{\alpha_1+ \cdots + \alpha_n} \leq \sum_{j=0}^\infty q^j $$
Take for example $n=5, z=\left(\frac 12, \frac 12, \frac 12, \frac 12, \frac 12\right), q=\frac 12 $, and the elements of $F$ are the 5 elements $\alpha_k \in \mathbb{N}^5$ with $|\alpha_k|=1$.
$$ \sum_{\alpha \in F} q^{\alpha_1+ \cdots + \alpha_n} = \frac 52 $$
$$ \sum_{j=0}^\infty q^j = \frac{1}{1-q} = 2 $$
I think how the proof should go from there is:
Given a natural number $j$, the number of elements $\alpha \in \mathbb{N}^n$ such that $|\alpha| = j$ is the number of ways to write $j$ as the sum of $n$ natural numbers $\alpha_1 + \cdots + \alpha_n = j$. This count is the same as the number of ways to write $j+n$ as the sum of $n$ positive integers. If we picture $j+n$ dots arranged in a line, this count is the same as the number of ways to choose $n-1$ of the $j+n-1$ positions in between dots as places to divide the dots into groups. So
$$ \big| \{ \alpha \in \mathbb{N}^n : |\alpha| = j\} \big| = {{n+j-1} \choose {n-1}} = {{n+j-1} \choose j} $$
$$ \sum_{\alpha \in F} q^{\alpha_1+\cdots+\alpha_n} \leq \sum_{\alpha \in \mathbb{N}^n} q^{|\alpha|} = \sum_{j=0}^\infty {{n+j-1} \choose j} q^j = \frac{1}{(1-q)^n} $$
so the sum $\sum_{\alpha \in F} |z^\alpha|$ converges at each $z \in P_1^n(0)$.
