linearly independence of $e^{a_1x},... e^{a_nx}$ 
$a_1,\ldots,a_n$ are real different numbers. Prove that the functions $e^{a_1x},...,e^{a_nx}$ are linearly independent in $Fun(R,R)$.

My way to try to prove it: 
I assumed: $b_1e^{a_1x} + \cdots + b_n e^{a_nx} = 0$, and we want to show that $b_1 = \cdots = b_n = 0$.
I differentiated it $n-1$ times, so I wrote it in $n\times n$ matrix. 
Then if I do transpose and calculate the determinant, I get something like: $b_1 \cdots b_n \times\text{Vandermonde Det}$.
I know Vandermonde det isn't $0$, but if one of the $b_i$'s is zero everything is zero and the det is $0$ so it is singular and it is linearly dependent! so I don't know why it is true at all.
 A: Hint: Look at the Wikipedia article on the Vandermonde matrix.
If the $a_i$ are distinct, then the matrix you get by looking at the various derivatives of the $e^{a_i x}$ at $x=0$ will have non-zero determinant.
This proves linear independence.
Start with the $0$-th derivatives. The first column of the matrix you will get is all $1$'s. The second column is $a_1, a_2$, and so on. The third column is obtained by differentiating twice, we get the $a_i^2$. 
I expect you have met the word if you have been assigned this exercise, but you may want to look up Wronskian.
Added: If one of the $a_i=0$, the above procedure doesn't quite work. Multiply all the functions by some $e^{cx}$, where $c+a_i\ne 0$ for any $a_i$. It is easy to see that the new set is linearly  independent if and only if the old set is, and now the Wronskian works smoothly. 
Or else note that the Wronskian is no longer quite of the Vandermonde type. However, it has a $1$ in the upper left corner, and $0$'s down the first column. So the Wronskian is $1$ times an $(n-1)\times (n-1)$ close relative of a Vandermonde matrix, whose determinant is non-zero. (The two fixes are essentially equivalent.)
A: Suppose that there exist $b_1,\dots,b_n$ such that for all $x \in \mathbb{R}$, $$b_1e^{a_1x}+ b_2e^{a_2x}+ \dots+ b_ne^{a_nx}=0$$
Then, derivating and evaluating at $0$, $$ b_1+ b_2+ \dots +b_n= 0 \\ b_1a_1+ b_2a_2+ \dots + a_nb_n=0 \\ b_1a_1^2+b_2a_2^2+ \dots + b_na_n^2=0 \\ \vdots \\ b_1a_1^{n-1}+b_2a_2^{n-1} + \dots + b_na_n^{n-1} =0$$
Therefore, the vectors $\left( \begin{matrix} 1 \\ a_1 \\ a_1^2 \\ \vdots \\ a_1^{n-1} \end{matrix} \right)$, $\left( \begin{matrix} 1 \\ a_2 \\ a_2^2 \\ \vdots \\ a_2^{n-1} \end{matrix} \right)$, ..., $\left( \begin{matrix} 1 \\ a_n \\ a_n^2 \\ \vdots \\ a_n^{n-1} \end{matrix} \right)$ are linearly dependent, hence:
$$0= \left| \begin{matrix} 1 & 1 &  \dots & 1 \\ a_1 & a_2 & \dots & a_n \\ a_1^2 & a_2^2 & \dots & a_n^2 \\ \vdots & \vdots & & \vdots \\ a_1^{n-1} & a_2^{n-1} & \dots & a_n^{n-1} \end{matrix} \right|$$
In this way, there is no $b_i$, and you deduce from Vandermonde's determinant that $a_i=a_j$ for some $i \neq j$.

Another method: Suppose that there exist $b_1,\dots,b_n$ such that for all $x \in \mathbb{R}$, $$b_1e^{a_1x}+b_2e^{a_2x}+\dots+b_ne^{a_nx}=0$$
Let $1 \leq k \leq n$ such that $a_k= \max\limits_{1 \leq i \leq n} a_i$. Then $$b_1e^{(a_1-a_k)x}+b_2e^{(a_2-a_k)x} + \dots + b_ne^{(a_n-a_k)x}=0$$
When $x \to + \infty$, you get $b_k=0$. So you can conclude by induction.
