Let $a\_1,\ldots,a\_n$ be $\mathbb Q$-linearly independant algebraic numbers. Are the functions $e^{az},\ldots,e^{a\_nz}$ algebraically independent functions (over $\mathbb C(z)$ or $\mathbb Q(z)$)?


I ask this because I wonder whether the Lindemann-Weierstrass theorem is a consequence of the Siegel-Shidlovskii theorem.


Thanks in advance for any hints.



algebraic independence of exponential functions
This can be checked by computing [the Wronskian](https://en.wikipedia.org/wiki/Wronskian) which results in the determinantal valuation
$$e^{(a\_1+\cdots+a\_n)z}V(a\_1,\dots,a\_n)$$
where $V:=V(a\_1,\dots,a\_n)$ is the determinant of the [Vandermonde matrix](https://en.wikipedia.org/wiki/Vandermonde_matrix)
$$V=\prod\_{1\leq i<j\leq n}(a\_j-a\_i).$$
So, it can only vanish when there are duplicates $a\_i=a\_j$ for some $i,j$. There is no need to have algebraic independence condition on the numbers. If the $a\_i$'s are distinct then the Wronskian does not vanish identically, hence the functions are linearly independent.


