I'm reading Transcendental number theory (Baker A) and now going through Lindemann's theorem. In the proof, he claims that
$$l^{np}(p-1)!\prod_{\substack{k=1\\k\ne i}}^n(\alpha_{i}-\alpha_{k})^{p}$$
is an algebraic integer divisible by $(p- 1)!$ but not by $p!$ if $p$ sufficiently large (p is prime), where $l$ is integer such that $l\alpha$'s are algebraic integer and $\{\alpha_1,...,\alpha_n\}$ is complete set of conjugates.
I had tried using minimal polynomial but it seems not work because the algebraic integer varies with p. Can anyone give details about this?
Given that $\beta_i:=l\alpha_i$ is an algebraic integer for all $i$, we can rewrite the expression to get $$l^{np}(p-1)!\prod_{\substack{k=1\\k\ne i}}^n(\alpha_{i}-\alpha_{k})^{p} =l^p(p-1)!\prod_{\substack{k=1\\k\ne i}}^n(\beta_{i}-\beta_{k})^{p},$$ which is clearly divisible by $(p-1)!$. Now for it not to be a multiple of $p!$ we need that the remaining factor $$l^p\prod_{\substack{k=1\\k\ne i}}^n(\beta_{i}-\beta_{k})^{p},$$ is not a multiple of $p$. Note that the discriminant of the number ring $\Bbb{Z}[\beta_1,\ldots,\beta_n]$ is given by $$\Delta:=\Delta(\Bbb{Z}[\beta_1,\ldots,\beta_n])=\prod_{i=1}^n\prod_{\substack{k=1\\k\ne i}}^n(\beta_{i}-\beta_{k}),$$ which is an integer (assuming that the $\beta_i$ are a complete set of conjugates over $\Bbb{Z}$). This shows that the remaining factor above divides the integer $(l\Delta)^p$, and this is clearly not a multiple of $p$ for sufficiently large $p$.