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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.

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.

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 $$C(z_1,\ldots,z_{n-1})\prod (z_n-z_1)\ldots(z_n-z_{n-1}).$$ 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. Vandermonde determinant is computed in ANY undergraduate textbook of linear algebra, as a first example of determinant.

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.

There are two standard proofs of this fact. Let $f_1,f_2,\ldots,f_n$ be either

a) analytic functions on some interval, OR

b) solutions of a linear differential equation $$y^{(n)}+a_{n-1}y^{(n-1)}+\ldots+a_0y=0$$ on some interval (with continuous functions $a_j$). Then $y_1, \ldots,y_n$ are linearly dependent if and only if their Wronskian determinant $$\left|\begin{array}{cccc}y_1&y_2&\ldots& y_n\\ y_1^\prime& y_2^\prime&\ldots&y_n^\prime\\ \ldots&\ldots&\ldots&\ldots\\ y_1^{(n-1)}& y_2^{(n-1)}&\ldots& y_n^{(n-1)}\end{array}\right|$$ is identically equal to $0$.

In our case, both assumptions hold: exponentials are analytic on any interval, and they also satisfy a differential equation with constant coefficients. The Wrosnkian determinant in question is the Vandermonde determinant and its computation is elementary.

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 under condition a) is problem 60. Wronskian criterion under condition b) is an immediate consequence of the uniqueness theorem for the linear differential equation.

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 $$C(z_1,\ldots,z_{n-1})\prod (z_n-z_1)\ldots(z_n-z_{n-1}).$$ 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. Vandermonde determinant is computed in ANY undergraduate textbook of linear algebra, as a first example of determinant.