Prove $e^x, e^{2x}..., e^{nx}$ is linear independent on the vector space of $\mathbb{R} \to \mathbb{R}$ Prove $e^x, e^{2x}..., e^{nx}$ is linear independent on the vector space of $\mathbb{R} \to \mathbb{R}$
isn't it suffice to say that $e^y$ for any $y \in \mathbb{R}$ is in $\mathbb{R}^+$
Therefore, there aren't $\gamma_1, ...\gamma_n$ such that $\gamma_1e^x+\gamma_2e^{2x}...+\gamma_ne^{nx}=0$.
Therefore, they're not linear dependent.  
I've seen a proof goes as follow:
take $(n-1)$ derivatives of the equation. then, you got $n$ equations with $n$ variables. Arranging it in a matrix (which is found out to be Van-Der-Monde matrix).
calculate the determinant which is $\ne 0$. Therefore, only the trivial solution exist. Therefore, no linear dependency.
Is all that necessary? 

The exercise is 
$$
f\alpha = \left\{
    \begin{array}{ll}
        \mathbb{R}\rightarrow\mathbb{R} \\
        \ t\mapsto e^{\alpha t}
    \end{array}
\right.
$$ Prove $(f_\alpha)_{\alpha \in\mathbb{R}} $is linear independent.
Let $(f_{\alpha_k})_{1\leq k \leq n} $ a finite number of vectors as $\alpha_1<\alpha_2<...<\alpha_n$.
When $\sum_{k=1}^n \alpha_k f_{\alpha_k}=0\Rightarrow \forall t\in\mathbb{R}, \sum_{k=1}^n \alpha_k e^{{\alpha_k} t}=0$
$$\Rightarrow \sum_{k=1}^{n-1} \alpha_k e^{{\alpha_k}t}=-\alpha_ne^{{\alpha_n}t}$$
$$ \Rightarrow\forall t\in\mathbb{R} \sum_{k=1}^{n-1} \alpha_k e^{({\alpha_k-\alpha_n})t}=-\alpha_n$$ where ${\alpha_k-\alpha_n}<0$ for $1\leq k \leq n-1$
$$\Rightarrow 0=-\alpha_n$$ when $n\rightarrow +\infty$
By repeating this process we obtain $\alpha_1=\alpha_2=...=\alpha_n=0$. QED
A better proof is using the fact that Eigenvectors with distinct eigenvalues are linearly independent 
Indeed,
$$f\alpha = \left\{
    \begin{array}{ll}
        \mathbb{R}\rightarrow\mathbb{R} \\
        \ t\mapsto e^{\alpha t}
    \end{array}
\right.
$$ is linearly independent as vectors are eigenvectors with distinct eigenvalues of:
$$D = \left\{
    \begin{array}{ll}
        C^\infty(\mathbb{R})\rightarrow C^\infty(\mathbb{R}) \\
        \ f\mapsto f'
    \end{array}
\right.$$
