Every divergence-free vector field generated from skew-symmetric matrix

Let $[a_{i,j}(x_1,\ldots,x_n)]$ be a skew-symmetric $n\times n$ matrix of functions $a_{i,j}\in C^\infty(\mathbb{R}^n)$. The vector field $$v=\sum\left(\dfrac{\partial}{\partial x_i}a_{i,j}\right)\dfrac{\partial}{\partial x_j}$$is divergence-free.

Prove by induction that for every $n\geq 2$, every $C^\infty$ divergence-free vector field on $\mathbb{R}^n$ is of this form.

Consider $n=2$. Suppose the vector field is $f_1(x_1,x_2)\dfrac{\partial}{\partial x_1}+f_2(x_1,x_2)\dfrac{\partial}{\partial x_2}$. Since the vector field is divergence-free, we have that $\dfrac{\partial}{\partial x_1}f_1(x_1,x_2)+\dfrac{\partial}{\partial x_2}f_2(x_1,x_2)=0$. By this result, there exists a function $g(x_1,x_2)$ whose $x_1$-derivative equals $f_2$ and whose $x_2$-derivative equals $-f_1$. The result follows.

But how about for $n>2$? To use induction, I have to relate a divergence-free vector field of $\mathbb{R}^n$ to a divergence-free vector field of $\mathbb{R}^{n-1}$. It is possible that the following result will help:

Let $v$ be a vector field on $\mathbb{R}^n$. Show that $v$ can be written as a sum $v=f_1\dfrac{\partial}{\partial x_1}+w$ where $w$ is a divergence-free vector field.