Determinant of matrix with entries $a_{ij}=e^{a_ib_j}$ 1) Let $a_1<\dots<a_n$ real numbers and $\lambda_1,\dots,\lambda_n\in\mathbb{R}\backslash\{0\}$
Let $f(x)=\lambda_1e^{a_1x}+\dots+\lambda_ne^{a_nx}$
Show that $f$ has at most $n-1$ zeroes
2) Furthermore, let $b_1<\dots<b_n$ real numbers.
Show that $\det A>0$ where $A$ is the matrix with coefficients $a_{ij}=e^{a_ib_j}$

For 1) : I tried a recurrence but gave up quite fast as it didn't seem to lead me anywhere. Then, I remarked that $\sum_{k=1}^n\lambda_ke^{a_kx}=0$ looked like what we do when trying to show that a set is linearly independent, but as that didn't hold for every $\lambda_k\in\mathbb{R}$ I was stuck.
For 2) : After searching, I didnt find anything good. I truly have no idea for this one.
 A: First Part
This is true for $n=1$.
Suppose this is true for $n-1$.
Since $\lambda_1e^{a_1x}\ne0$ we can divide by that to get
$$
1+\frac{\lambda_2}{\lambda_1}e^{(a_2-a_1)x}+\frac{\lambda_3}{\lambda_1}e^{(a_3-a_1)x}+\dots+\frac{\lambda_n}{\lambda_1}e^{(a_n-a_1)x}
$$
taking the derivative gives
$$
\frac{\lambda_2}{\lambda_1}(a_2-a_1)e^{(a_2-a_1)x}+\frac{\lambda_3}{\lambda_1}(a_3-a_1)e^{(a_3-a_1)x}+\dots+\frac{\lambda_n}{\lambda_1}(a_n-a_1)e^{(a_n-a_1)x}
$$
which, according to the inductive hypothesis, has only $n-2$ zeros. If the derivative of a function has $n-2$ zeros, the function can only have $n-1$ zeros.

Second Part
Note that the first part is true as long as at least one of the $\lambda_k$ is not zero. If at least $m$ ($1\le m\le n$) of the $\lambda_k$ are not zero, we can use the result for $m$ instead and get that there are at most $m-1\le n-1$ zeros.
Suppose that
$$
A^T\begin{bmatrix}\lambda_1\\\lambda_2\\\vdots\\\lambda_n\end{bmatrix}
=\begin{bmatrix}f(b_1)\\f(b_2)\\\vdots\\f(b_n)\end{bmatrix}=0
$$
Since $f$ has at most $n-1$ zeros, all of the $\lambda_k$ must be $0$. Therefore, $A$ is not singular, and thus $\det(A)\ne0$.
