Each $w\in W$ can be expressed as product of distinct reflections? For a Weyl group $W$, I would like to know whether each $w\in W$ can be expressed as $w=s_{\alpha_1}s_{\alpha_2}\cdots s_{\alpha_k}$ for some distinct positive roots $\{\alpha_1, \alpha_2, \cdots, \alpha_k\}\subseteq \Phi^+$.
I know for type A, the above is true. 
Since $W(A_n)\cong S_{n+1}$ with the map $s_{e_i-e_j}\mapsto (i \ j)$. For $w\in W(A_n)$, let $\sigma\in S_{n+1}$ be the image of $w$ under the map. 
By the fact that every permutation can be decomposed as a product of disjoint cycles in the form of $(a_1\cdots a_l)$. 
    And also the fact that $(a_1\cdots a_l)=(a_1 \ a_l)\cdots (a_1 \ a_2)$. We get $\sigma$ is a product of distinct transpositions. And distinct transpositions correspond to distinct reflections of the form $s_{e_i-e_j}$. Therefore, the claim follows.  
I would like to know whether the above statement is also true for other types?
 A: The answer is "yes" and there is a geometric explanation.
Let $\mathcal{H}$ denote the set of hyperplanes corresponding to the reflections $s_\alpha$ with $\alpha\in\Phi^+$ (note that $s_\alpha=s_{-\alpha}$), and let $\Sigma$ denote the connected components of $V\setminus\bigcup_{H\in\mathcal{H}}H$ (where $V$ is the vector space where $\Phi$ lives). The elements of $\Sigma$ are the chambers of the Coxeter complex of $(V,\Phi)$, and it is well-known that $W$ acts simply transitively on $\Sigma$.
Fix some $C_0\in\Sigma$, let $w\in W$, and put $C_1=w(C_0)$.
Then you can "walk" from the chamber $C_0$ to $C_1$ by crossing through the walls of the chambers (here, a wall means a codimension-$1$ face of a chamber). Every time you cross from $C$ into an adjacent chamber $C'$, you walk though some hyperplane $H\in \mathcal{H}$, and in that case you have $C'=s_\alpha C$, where $\alpha\in \Phi^+$ corresponds to $H$.
Since you can walk from $C_0$ to $C_1=wC_0$ without crossing the same hyperplane $H$ twice (Edit: e.g. take a walk minimizing the number of chambers visited, or go along a straight line in general position w.r.t. $\mathcal{H}$), it follows that $C_1=s_{\alpha_t}\cdots s_{\alpha_1}C_0$, where $\alpha_t,\dots,\alpha_1\in\Phi^+$ are distinct and correspond to the hyperplanes you crossed on your way from $C_0$ to $C_1$.
Thus, $w=s_{\alpha_t}\cdots s_{\alpha_1}$ with distinct $\alpha_t,\dots,\alpha_1\in \Phi^+$.
A: After reading your comments, I come up with the following proof. I would like to know whether my proof is correct or not.
For weyl group $W$, each $w\in W$ can be expressed as $w=s_{\beta_l}\cdots s_{\beta_{2}} s_{\beta_1}$ for some distinct positive roots $\{\beta_1, \beta_2, \cdots, \beta_l\}\subseteq \Phi^+$.
My proof:
Let $w=s_{i_1}s_{i_2}\cdots s_{i_l}$ be a reduced expression with $s_{i_k}$ is the simple reflection of root $\alpha_{i_k}\in\Delta$.
Let $w_k=s_{i_1}s_{i_2}\cdots s_{i_k}$, $w_0=1$, $C_k=w_{k}C_+$ and $C_0=C_+$. By direct computation,
$
C_k=w_{k}C_+
=w_{k-1}s_{i_k}C_+
=w_{k-1}s_{i_k}w_{k-1}^{-1}w_{k-1}C_+
=w_{k-1}s_{i_k}w_{k-1}^{-1}C_{k-1}
=s_{\beta_k}C_{k-1},
$
where $\beta_k:=w_{k-1}\alpha_{i_k}=s_{i_1}s_{i_2}\cdots s_{i_{k-1}}(\alpha_{i_k})$ and $\beta_1=\alpha_{i_1}$.
Note that 
$
wC_+
=C_l
=s_{\beta_l}C_{l-1}
=s_{\beta_l}s_{\beta_{l-1}}C_{l-2}=\cdots
=s_{\beta_l}s_{\beta_{l-1}}\cdots s_{\beta_2}C_1
=s_{\beta_l}s_{\beta_{l-1}}\cdots s_{\beta_1}C_+.
$
By the simply-transitivity of $W$, we get $w=s_{\beta_l}s_{\beta_{l-1}}\cdots s_{\beta_1}$.
Since $s_\alpha=s_{-\alpha}$ for all $\alpha\in\Phi^+$, we may WLOG assume $\beta_k\in\Phi^+$.
Suppose $\beta_k=\beta_j$ for some $j<k$, then $s_{\beta_k}=s_{\beta_j}$.
Then 
$
w_{k-1}s_{i_k}w_{k-1}^{-1}=w_{j-1}s_{i_j}w_{j-1}^{-1}.
$
$
s_{i_j}=(s_{i_j}\cdots s_{i_{k-1}})s_{i_k}(s_{i_j}\cdots s_{i_{k-1}})^{-1},
$
from which we obtain, upon right-multiplying by $s_{i_j}\cdots s_{i_k}$.
$
s_{i_{j+1}}\cdots s_{i_{k-1}}
=s_{i_{j}}\cdots s_{i_{k}}.
$
Then
$
w=s_{i_1}\cdots s_{i_l}=s_{i_1}\cdots\hat{s_{i_j}}\cdots\hat{s_{i_k}}\cdots s_{i_l},
$
a contradiction to the fact that $w=s_{i_1}s_{i_2}\cdots s_{i_l}$ is a reduced expression.
Therefore, the claim follows.
