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? 
 A: 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:
$$D = \left\{
    \begin{array}{ll}
        C^\infty(\mathbb{R})\rightarrow C^\infty(\mathbb{R}) \\
        \ f\mapsto f'
    \end{array}
\right.$$
A: 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.
A: To show that the functions $e^{\alpha_i x}$ are linearly independent over $\Bbb R$ for distinct, real but otherwise arbitrary $\alpha_i$, then the arguments presented by Davide Giraudo and julien in their answers seem to be the way to go.  So, +1 for each of their answers and lauds for their fine work!
However, if one is primarily interested in showing the functions $e^x$, $e^{2x}$, . . . $e^{nx}$, etc., are linearly independent, that is, the functions $e^{mx}$ for positive integral $m$, as appears to be indicated in the title and body of the question as stated, then the following little trick may be of interest:  suppose there existed $\beta _i \in \Bbb R$ with 
$\sum_1^n \beta_i e^{ix} = 0; \tag{1}$
then, since $e^{ix} = (e^x)^i$, (1) becomes
$\sum_1^n \beta_i (e^x)^i = 0, \tag{2}$
that is, $e^x$ must be a (real) zero of the polynomial equation
$p(y) = \sum_1^n \beta_i y^i = 0, \tag{3}$
which implies that $e^x$ can only take on at most $n$ values which will be among the real zeroes of $p(y)$.  And that just won't work, will it?
Note Added in Edit, Friday 10 January 2014 11:43 AM PST:  In the light of the warm reception, in terms of upvotes, this answer has received and also in the light of Christoph Pegel's comment, I would like to point out that this approach goes a long way toward completely algebraicizing this problem, at least for functions of the form $e^{ix}$ and related.  As indicated by Christoph Pegel, any function satisfying $f(mx) = (f(x))^m$ for positive integral $m$ will be susceptible to this argument, viz. we would have to have
$\sum_1^n \beta_i (f(x))^i = 0 \tag{4}$
if the $f(nx)$ were linearly dependent.  Note that the fact that (3) or (4) have at most finite number of zeroes is a purely algebraic result, depending only on the Euclidean division formula $p(x) = (x - \lambda)q(x)$ for $\lambda$ a root of $p(x)$; in fact, (4) shows the $(f(x))^i$ are linearly independent if $f(x)$ takes on an infinite number of values.  The results extend to the cases of other $f$, i.e. $f(x) = a^x$ for nonzero real $a$ etc., and apparently even to other base fields than $\Bbb R$.  A pretty algebraic situation, really, for the $e^{mx}$ and for any other $f$ such that $f(mx) = ((f(x))^m$.  End of Note.
Hope this helps!  Cheers,
and as always,
Fiat Lux!!!
A: Since the $\gamma_i$ (using your notation) can be negative, it does not suffice to state that $e^x > 0 \; \forall x\in\mathbb{R}$.
You can either use the matrix/determinant method, or, I believe, you can look at the power series of $e^x$ to do so.
A: let $e=1$. Then if the values are linearly dependent, $e$ would be a root of a polynomial.  EDIT: I just realized this argument is wrong, because the coefficients of the polynomial are from $\mathbb R$.
A: You can show it using the wronskian:
Let $f_i=e^{ix}$
Then,
$$W(f_1,f_2,...,f_n)=
\left| \begin{array}{ccc}
e^x & e^{2x} & ... & e^{nx}\\
e^x & 2e^{2x} & ...& ne^{nx} \\
e^x & 4e^{2x} &... & n^2e^{nx}\\
...&...&...&...\\
e^x&2^{n-1}e^{2x} & ... & n^{n-1}e^{nx}\end{array} \right| $$
Dividing the ith colunm by $e^{ix}>0$ gives:
$$
\left| \begin{array}{ccc}
1 & 1 & ... & 1\\
1 & 2 & ...& n \\
1 & 4 &... & n^2\\
...&...&...&...\\
1 &2^{n-1} & ... & n^{n-1}\end{array} \right| $$
Which suprisingly seem to be $0!*1!*2!*...*(n-1)!$, but I don't know yet how to prove it.
EDIT:
As I was anwered, this is Vandermonde matrix, which known to be with positive determinant.
