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? 

Positiveness of the exponential is not enough as pointed out in the comments and anorton's answer. 
Start from the equation $$\forall x\in\mathbb R, \quad \sum_{j=1}^n\gamma_je^{jx}=0.$$
Multiply this equation by $e^{-nx}$. We get for any $x$,
$$\gamma_n+\sum_{j=1}^{n-1}\gamma_je^{(j-n)x}=0.$$
Now, letting $x\to+\infty$, we obtain $\gamma_n=0$. We can either repeat the procedure or write it properly by induction.
