You can use the Vandermonde determinant.
Let $$\beta_1e^{-\alpha_1x} + \cdots + \beta_ne^{-\alpha_1x} = 0$$
Plugging in $x = 0, 1, \ldots, n-1$ yields:
$$\beta_1 + \cdots + \beta_n = 0$$
$$\beta_1e^{-\alpha_1} + \cdots + \beta_ne^{-\alpha_n} = 0$$
$$\beta_1e^{-2\alpha_1} + \cdots + \beta_ne^{-2\alpha_n} = 0$$
$$\vdots$$
$$\beta_1e^{-(n-1)\alpha_1} + \cdots + \beta_ne^{-(n-1)\alpha_n} = 0$$
The determinant of this linear system is the Vandermonde determinant:
$$\begin{vmatrix}
1 & 1 & \cdots & 1\\
e^{-\alpha_1} & e^{-\alpha_2} & \cdots & e^{-\alpha_n}\\
e^{-2\alpha_1} & e^{-2\alpha_2} & \cdots & e^{-2\alpha_n}\\
\vdots & \vdots & \ddots & \vdots\\
e^{-(n-1)\alpha_1} & e^{-(n-1)\alpha_2} & \cdots & e^{-(n-1)\alpha_n}\\
\end{vmatrix} = \prod_{1 \le i < j \le n} (e^{-\alpha_j} - e^{-\alpha_i}) \ne 0$$
because $e^{-\alpha_i} \ne e^{-\alpha_j}$ for all $i \ne j$. Hence, the system has the unique solution $\beta_1= \beta_2 = \cdots = \beta_n = 0$, which implies linear independence.
Another solution using Vandermonde.
Assume
$$\beta_1e^{-\alpha_1x} + \cdots + \beta_ne^{-\alpha_1x} = 0$$
Taking the derivative $n-1$ times yields:
$$\beta_1e^{-\alpha_1x} + \cdots + \beta_ne^{-\alpha_nx} = 0$$
$$-\alpha_1\beta_1e^{-\alpha_1x} - \cdots - \alpha_n\beta_ne^{-\alpha_nx} = 0$$
$$\alpha_1^2\beta_1e^{-\alpha_1x} + \cdots + \alpha_n^2\beta_ne^{-\alpha_nx} = 0$$
$$\vdots$$
$$\alpha_1^{n-1}(-1)^{n-1}\beta_1e^{-\alpha_1x} + \cdots - \alpha_n^{n-1}(-1)^{n-1}\beta_ne^{-\alpha_nx} = 0$$
The determinant of this linear system is again the Vandermonde determinant:
$$\begin{vmatrix}
e^{-\alpha_1x} & e^{-\alpha_2x} & \cdots & e^{-\alpha_nx}\\
(-\alpha_1)e^{-\alpha_1x} & (-\alpha_2)e^{-\alpha_2x} & \cdots & (-\alpha_n)e^{-\alpha_nx}\\
(-\alpha_1)^2e^{-\alpha_1x} & (-\alpha_2)^2e^{-\alpha_2x} & \cdots & (-\alpha_n)e^{-\alpha_nx}\\
\vdots & \vdots & \ddots & \vdots\\
(-\alpha_1)^{n-1}e^{-\alpha_1x} & (-\alpha_2)^{n-1}e^{-\alpha_2x} & \cdots & (-\alpha_n)^{n-1}e^{-\alpha_nx}\\
\end{vmatrix}$$
$$ = e^{-\alpha_1x}\cdots e^{-\alpha_nx}
\begin{vmatrix}
1 & 1 & \cdots & 1\\
-\alpha_1 & -\alpha_2 & \cdots & -\alpha_n\\
(-\alpha_1)^2 & (-\alpha_2)^2 & \cdots & (-\alpha_n)^2\\
\vdots & \vdots & \ddots & \vdots\\
(-\alpha_1)^{n-1} & (-\alpha_2)^{n-1} & \cdots & (-\alpha_n)^{n-1}\\
\end{vmatrix} =e^{-\alpha_1x}\cdots e^{-\alpha_nx} \prod_{1 \le i < j \le n} (\alpha_i - \alpha_j) \ne 0$$
because $\alpha_i \ne \alpha_j$ for all $i \ne j$. Hence, the system has the unique solution $\beta_1= \beta_2 = \cdots = \beta_n = 0$, which implies linear independence.