I don't understand part of a proof of the Lindemann-Weierstrass theorem I'm looking at this PlanetMath article on the Lindemann-Weierstrass theorem, specifically the dicussion right after equation (4). I think this is about the same as what appears below equation (5) in Baker's text, p6. There is a statement in there that I can't quite understand.
The article defines $\alpha_1,...,\alpha_N$ as the roots of an irreducible polynomial with integer coefficients, and integers $\beta_1,...,\beta_N$. It defines $S_N$ as the set of permutations of $N$ elements. It then considers the expansion.
$$ \prod_{\sigma \in S_N} (\beta_1 e^{\alpha_{\sigma(1)}} + ... + \beta_N e^{\alpha_{\sigma(N)}})$$
It notes that "[t]here are $N!$ factors in this product, so expanding the product, it is a sum of terms of the form $e^{h_1\alpha_1 + ... h_N \alpha_N}$ with integral coefficients, and
$h_1+...+h_N=N!$. The set of all such exponents forms a complete set of conjugates." I'm good so far.
Then it says that "[b]y symmetry considerations, we see that the coefficients of two conjugate terms are equal."
This is the part that I don't get. Any explanation of this is welcome, but the rest of the discussion might help you to see where my confusion is coming from.
It seems to me that, for each list of nonnegative integer h-values such that $h_1+⋯+h_N=N!$, the set of terms whose exponent has coefficients in this list (in some order) has two properties: a) each term has the same coefficient, b) the associated set of exponents, "$R$," is the set of roots of some rational polynomial, because $\prod_{r \in R} (x-r)$ is symmetric in the $\alpha$ values and the $\alpha$ values are a complete set of conjugates.
If that's true, I think it proves the bold statement above, but I'm not sure. As far as I can tell, the same exponent might appear multiple times, along with its conjugates, associated with different partitions of $N!$ into $h_1,...,h_N$. In this case I think I would need to collect terms together in order to show the truth of the bold statement.
 A: Maybe it is easier if we remove some of the notation. Consider
$$F(a_1, ..., a_N) = \prod_{\sigma \in S_N} (\beta_1 a_{\sigma(1)} + ... + \beta_Na_{\sigma(N)})$$
and we want to know that if $\tau \in S_N$ (more precisely the galois group embedded in $S_N$) then the terms $a_1^{h_1}...a_N^{h_N}$ and $a_{\tau(1)}^{h_1}...a_{\tau(N)}^{h_N}$ have the same coefficients.
But the above polynomial $F$ is symmetric, that is, $F(a_{\tau(1)}, ..., a_{\tau(N)}) = F(a_{1}, ..., a_{N})$. This is becuase
\begin{align*}
F(a_{\tau(1)}, ..., a_{\tau(N)}) &= \prod_{\sigma \in S_N} (\beta_1 a_{\sigma(\tau(1))} + ... + \beta_N a_{\sigma(\tau(N))}) \\
&= \prod_{\gamma = \sigma\tau \; : \; \sigma \in S_N} (\beta_1 a_{\gamma(1)} + ... + \beta_N a_{\gamma(N)}) \\
&= \prod_{\gamma \in S_N} (\beta_1 a_{\gamma(1)} + ... + \beta_N a_{\gamma(N)}) \\
&= F(a_1, ..., a_N)
\end{align*}
Hence the monomials $a_1^{h_1}...a_N^{h_N}$ and $a_{\tau(1)}^{h_1}...a_{\tau(N)}^{h_N}$ have the same coefficients.
In our case we may take $a_i = e^{\alpha_i}$.
