I am studying the proof of the simple version of Lindemann's Theorem.
Theorem. If $\alpha$ is a non-zero algebraic number, then $e^\alpha$ is transcendental.
Apologies for the long read, but to explain my question I need to give a very brief summary of the first part of the proof.
Let $\alpha$ be algebraic with conjugates $\alpha_1,\ldots,\alpha_l$. Suppose $e^\alpha$ is algebraic with minimal polynomial $$p(z)=c_0+c_1z+\cdots+c_mz^m$$ over the integers. We expand $$p(e^{\alpha_1})\cdots p(e^{\alpha_l})\tag{1}$$ as a sum $$\sum_\beta a_\beta e^\beta\ ,$$ where each $\beta$ is an integer linear combination of the $\alpha_j$. By collecting terms we obtain a similar sum in which all the $\beta$ are distinct. Now it is clear that one of the $\beta$ values is zero, so that we have an integer term $c_0^l$. However it is possible that more than one $\beta$ is zero, so after collecting terms we have something like $$a_0+a_1e^{\beta_1}+\cdots\ ,\tag{2}$$ where the $\beta_j$ are non-zero. If $a_0\ne0$ then the proof proceeds quite smoothly, but if $a_0=0$ then we have to modify the preceding expression... but I don't need to tell you about that.
My question: is this modification ever actually necessary? That is, can $a_0$ ever be zero? To be quite clear about it -
Can anyone provide a specific example of an algebraic number $\alpha$ and an irreducible polynomial $p$ such that when we expand $(1)$ we obtain an expression like $(2)$ with $a_0=0$?
Possible example (some details to work out): Let $p(z)=144z^5-27z^3-110z^2-84z-133$ which I expect to be irreducible over the rationals (should be a simple check left to the OP) and let $\alpha_1,..\alpha_5$ the roots.
Then since the sum of the roots is zero (and again I expect no other integral linear relation between them though it is something to be checked), when one does $p(e^{\alpha_1})\cdots p(e^{\alpha_5})$, one gets a free term which is $144^5-27^5-110^5-84^5-133^5=0$ per the famous Lander and Parkin counterexample to Euler's conjecture