Linear independence of exponential functions: a reference

I will recount the more general statement of linear independence of characters, given in Lang's Algebra book, and credited to Artin. Let $G$ be a group, and $K$ a field. Then distinct homomorphisms $\phi_1, \ldots, \phi_n: G \to K^\times$ are linearly independent.

Proof: Suppose not, and suppose we have a nontrivial linear relation

$$a_1 \phi_1 + \ldots + a_n \phi_n = 0,\qquad (1)$$

where $n$ is taken as small as possible. Clearly $n>1$ and $a_i \neq 0$ for all $i$. Because the $\phi_i$ are distinct, we can find an element $g \in G$ such that $\phi_1(g) \neq \phi_2(g)$. We have

$$a_1 \phi_1(gh) + a_2 \phi_2(gh) + \ldots + a_n \phi_n(gh) = 0$$

for all $h \in G$; by virtue of the $\phi_i$ being homomorphisms, this may be rewritten to say

$$a_1 \phi_1(g)\phi_1 + a_2 \phi_2(g)\phi_2 + \ldots + a_n \phi_n(g)\phi_n = 0, \qquad (2)$$

Dividing $(2)$ by $\phi_1(g)$ and then subtracting (1) from the result, we arrive at a linear relation

$$\left(a_2 \frac{\phi_2(g)}{\phi_1(g)} - a_2\right) \phi_2 + \ldots = 0$$

which has fewer than $n$ summands and is nontrivial by choice of $g$, contradiction. $\Box$

Let $y_k(t)=e^{tz_k}$. Proving by contradiction, suppose that they are linearly dependent, that is $$\sum_{k=1}^nc_ky_k\equiv 0.$$ Differentiating $n-1$ times we obtain a homogeneous system of linear equations with respect to $c_k$. To have a non-trivial solution, this system must have non-zero determinant. The determinant is: $$\left|\begin{array}{cccc}y_1&y_2&\ldots& y_n\\ y_1^\prime& y_2^\prime&\ldots&y_n^\prime\\ \ldots&\ldots&\ldots&\ldots\\ y_1^{(n-1)}& y_2^{(n-1)}&\ldots& y_n^{(n-1)}\end{array}\right|=A(t) \left|\begin{array}{cccc}1&1&\ldots&1\\ z_1&z_2&\ldots& z_n\\ \ldots&\ldots&\ldots&\ldots\\ z_1^{n-1}&z_2^{n-1}&\ldots&z_n^{n-1}\end{array}\right|,$$ where $A(t)=e^{t(z_1+\ldots+z_n)}\neq 0$. The determinant in the right hand side is easy to compute. Consider it as a polynomial with respect to, $z_n$. It is evidently of degree $n-1$ and has $n-1$ roots at $z_1,\ldots,z_{n-1}$. Therefore it is of the corm $$C(z_1,\ldots,z_{n-1})(z_n-z_1)\ldots(z_n-z_{n-1}).$$ Looking at the top degree term, we conclude that $C$ is a similar polynomial. So by induction our determinant is $$\prod_{i<k}(z_i-z_k).$$ this is never zero, since $z_k$ are distinct.

References. Polya Szego, Problems and theorems of analysis, vol II, Part 7, "Determinants and quadratic forms''. Computation of the Vandermonde determinant is problem 2. The Wronskian criterion of linear independence is problem 60.

Remark. Vandermondes's determinant is computed in ANY undergraduate textbook of linear algebra, as a first example of determinant. For example, I teach linear algebra with the textbook of Strang, and differential equations with the textbook of Boyce and di Prima. Both of them have Vandermonde determinant.

Remark 2. Undergraduate textbooks are rarely freely available online. If you insist on a free online reference, you may refer on the proof above.