Explanation of proof of the transendence of $\alpha_1^{\beta_1} \dotsm \alpha_n^{\beta_n}$

Question

I'm following this proof in the book "Transcendental Numbers" by M. Ram Murty and Purusottam Rath.

The result is a corollary of Baker's theorem.

There are a couple of things I don't understand. Why is it sufficient to show that $$\beta_1\log\alpha_1+\ldots+\beta_m\log\alpha_m=0$$ for any $\alpha_i$ and $\beta_i$ satisfying the given conditions? I can see that $$\log(\alpha_1^{\beta_1}\dotsm\alpha_m^{\beta_m})$$ is equal to the left hand side of the above, but I'm not sure why/how this is useful, or where the contradiction comes in.

Secondly, I understand that the linear independence of $\beta_1, \dotsc, \beta_m$ implies that $$c_mA_1=\dotso=c_mA_{m-1}=A_1c_1+\dotsb+A_{m-1}c_{m-1}=0$$ But how does this imply that $A_1=\dotso=A_{m-1}=0$?

Any help would be appreciated.

Edit: added missing "$=0$" on the last line

Answer 1

1. Suppose $\alpha_1^{\beta_1}\cdots\alpha_m^{\beta_m}$ is algebraic. Call it $\alpha_{m+1}$; then $$\beta_1\log\alpha_1+\cdots+\beta_m\log\alpha_m+\beta_{m+1}\log\alpha_{m+1}=0$$ with $\beta_{m+1}=-1$ (as is written in the text), contrary to what's being shown.

2. It's supposed that $c_m\neq 0$ (see the text). Thus, $A_1=\ldots=A_{m-1}=0$.