How do one get the equation below in the proof in [Chow 1999]?
Chow writes on page 444 - 445:
"If $A=(\alpha_1,\alpha_2,...,\alpha_n)$ is a finite sequence of complex numbers, then for brevity we write $A_i$ for the field $\mathbb{Q}(\alpha_1,e^{\alpha_1},\alpha_2,e^{\alpha_2},...,\alpha_i,e^{\alpha_i})$. In particular, $A_0=\mathbb{Q}$.
Definition. A tower is a finite sequence $A=(\alpha_1,\alpha_2,...,\alpha_n)$ of nonzero complex numbers such that for all $i\in\{1,2,...,n\}$, there exists some integer $m_i>0$ such that $\alpha_i^{m_i}\in A_{i-1}$ or $e^{\alpha_im_i}\in A_{i-1}$ (or both). A tower is reduced if the set $\{\alpha_i\}$ is linearly independent over $\mathbb{Q}$. If $\beta\in\mathbb{C}$, then a tower for $\beta$ is a tower $A=(\alpha_1,\alpha_2,...,\alpha_n)$ such that $\beta\in A_n$.
...
... there is a reduced tower $A=(\alpha_1,\alpha_2,...,\alpha_n)$ for $R$.
...
But $A$ is reduced, so
$$R=\sum_{i=1}^n\frac{p_i\alpha_i}{q_i}$$
for some integers $p_1,q_1,p_2,q_2,...p_n,q_n$."
until now, I only see:
Because $A$ is reduced, $\{\alpha_1,\alpha_2,...,\alpha_n\}$ is linearly independent over $\mathbb{Q}$.
Because $A$ is a tower for $R$, $R$ is algebraically over $\mathbb{Q}(\alpha_1,e^{\alpha_1},\alpha_2,e^{\alpha_2},...,\alpha_n,e^{\alpha_n})$. So there is an algebraic relation for $R$ in dependence of $\alpha_1,e^{\alpha_1},\alpha_2,e^{\alpha_2},...,\alpha_n,e^{\alpha_n}$.
Where does the above linear equation come from?
$\ $
[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448
And that is the problem. The reasons $R$ must satisfy this linear equation are in the lines you didn't see - or at least left out here:
$R$ is not some arbitrary number. He is trying to prove that Schanuel's conjecture proves his "Conjecture 1": If $R + e^R = 0$, then $R\notin \Bbb E$. He is doing a proof by contradiction, so he assumes $R \in \Bbb E$.
After showing that there is some reduced tower $A = (\alpha_1, \alpha_2, \ldots,\alpha_n)$ with $R \in A_n$, he extends the tower by adding $R$ and calls it $A'$. So $A'_i = A_i$ for $i\le n$.
If $A'$ is also reduced, then by the first point above, either $R$ or $e^R$ is algebraic over $A'_n = A_n$, but not both. This cannot be, because $e^R = -R$, and so is algebraic over $A_n$ if and only if $R$ is.
So, we know that $A'$ is not a reduced tower. By definition, this means that $\alpha_1, \alpha_2, \ldots,\alpha_n, R$ are linearly dependent over $\Bbb Q$. That is, there are $\rho_i \in \Bbb Q$, not all zero, with $$\left(\sum_{i=1}^n \rho_i\alpha_i\right) + \rho_{n+1}R = 0$$
If $\rho_{n+1} = 0$, then $\sum_{i=1}^n \rho_i\alpha_i = 0$, making the $\alpha_i$ linearly dependent. But $A$ is reduced, so this cannot be. Thus $\rho_{n+1} \ne 0$ and $$R = \sum_{i=1}^n -\frac{\rho_i}{\rho_{n+1}}\alpha_i$$
Rewrite the rational numbers $-\frac{\rho_i}{\rho_{n+1}}$ as the ratio of integers $p_i$ and $q_i$, and you get the linear equation.