Prove that $\{\mathbb e^{r_1x}, \mathbb e^{r_2x}..., \mathbb e^{r_nx}\}$ is linear independent Prove that $\{\mathbb e^{r_1x}, \mathbb e^{r_2x}..., \mathbb e^{r_nx}\}$ are  linearly independent where $r_1,r_2,...r_n$ are not equal.
now I know that this question has been asked before but it's not the same this doesn't end up to be a  Vandermonde matrix.
so is there a workaround this as well.
 A: By re-indexing, we can assume $r_{1} > r_{2} > ... > r_{n}$. 
Suppose there are $c_{i}$ not all equal zero such that $\sum_{i=1}^{n}c_{i}e^{r_{i}x} =0$. Let $k$ be the smallest index such that $c_{k} \neq 0$. Then $\sum_{i=1}^{n}c_{i}e^{r_{i}x} =0 \implies \sum_{i=k}^{n}c_{i}e^{r_{i}x} =0$.
Then on multiplying both sides of this shorter sum by $e^{-r_{k}x}$, we get $\sum_{i=k}^{n}c_{i}e^{(r_{i} - r_{k})x} = 0$. Since the LHS is a continuous function, we take the limit as $x \to \infty$ and we get $c_{k} = 0$, which is a contradiction. 
A: Say $c_1.e^{r_1 x} + c_2.e^{r_2 x} + ... + c_n.e^{r_n x} = 0$
Differentiate w.r.t. x:
$r_1.c_1.e^{r_1 x} + r_2.c_2.e^{r_2 x} + ... + r_n.c_n.e^{r_n x} = 0$
n-th order derivative:
$r_1^n.c_1.e^{r_1 x} + r_2^n.c_2.e^{r_2 x} + ... + r_n^n.c_n.e^{r_n x} = 0$
Now we have 
$\begin{bmatrix}
1   & 1   &... & 1\\
r_1 & r_2 &... & r_n\\
.   & .   &... & .  \\
.   & .   &... & .  \\
r_1^n & r_2^n &... & r_n^n\end{bmatrix}$ . 
$\begin{bmatrix}
c_1.e^{r_1 x} \\
.   \\
.   \\
c_n.e^{r_n x} 
\end{bmatrix}$ = $0$.
The matrix is Vander Monde matrix, so it is invertible, thus 
$\begin{bmatrix}
c_1.e^{r_1 x} \\
.   \\
.   \\
c_n.e^{r_n x} \end{bmatrix} = 0$.
That is each $c_k.e^{r_k x} = 0$, which means $c_k = 0$.
