How to find a local chart of the Straightening Theorem My question is how to prove this result:
Let $X$ a smooth vector field on a smooth manifold $M$ and $p\in M$ such that $X(p)\neq0\Rightarrow$ 
$\forall\hspace{0.2cm}(U,\eta)$ local chart  centered at $p$, there exist a local chart $(U,\psi)$ centered at $p$ such that $X(p)=\frac{\partial}{\partial \psi_1}\vert_p$.
I'm thinking to this proof: with the first chart $\eta=(\eta_1,...,\eta_n)$ we could write $X(p)$ with the local (chart) coordinates:
$X(p)=\sum_{i=1...n}X^i(p) \frac{\partial}{\partial\eta_i} \vert _p\neq 0$, so there exist $1\leq j\leq n$ such that $X^j(p)\neq 0.$
Then I define $I=\{i\vert X^i(p)\neq0\}$, and $\psi=(\psi_1,...,\psi_n):=(\frac{1}{\vert I \vert}\sum_{i\in I}\frac{1}{X_i(p)}\eta_i,\frac{\eta^3_1}{3},...,\frac{\eta^3_{j-1}}{3},\frac{\eta^3_{j+1}}{3},...,\frac{\eta^3_n}{3})$.
Now, in the new local coordinates, $X(p)=\sum_{i=1...n}Y^i \frac{\partial}{\partial\psi_i} \vert _p$, and knowing that:
$(Y^1(p),...,Y^n(p))^T=\mathcal{J}ac(\psi\circ\eta^{-1})_{\eta(p)}\cdot(X^1(p),...,X^n(p))^T$, we have:
$\mathcal{J}ac(\psi\circ\eta^{-1})_{\eta(p)}=\begin{bmatrix}
\frac{1}{nX^1(p)} & \frac{1}{nX^2(p)} & ... &\frac{1}{nX^n(p)} \\
0 & ... & ... & 0\\
. & . & . & .\\
. & . & . & .\\
. & . & . & .\\
0 & ... & ... & 0
\end{bmatrix}$
Then $Y^1(p)=\sum_{i=1,...,n}\frac{1}{nX^i(p)}X^i(p)=1$ and $Y^i(p)=0\hspace{0.2cm}\forall\hspace{0.2cm}i=2,...,n\Rightarrow X(p)=\frac{\partial}{\partial \psi_1}\vert_p$.
(To obtain $0$ I've used the fact that $\eta^2_i(p)=0$ because the chart is centered at $p$, and in the first row I've assumed wlog that $X^i(p)\neq0 \hspace{0.2cm} \forall i$)
Is this proof ok? Is there any easyer way then the above one?
 A: The determinant of the Jacobian of your change of charts seems to be zero, so I don't think that proof works.
One way to prove the result is the following: Since $X(p) \neq 0$, there exist $\omega_1,\cdots,\omega_n \in T_pM^*$ such that they are linearly independent and $\omega_1(X(p))=1$, $\omega_i(X(p))=0$ for every $i \neq 1$. By making $\omega_i(x)=\omega_i$ in the chart $U$, we have $d\omega_i=0$. Picking a chart which is homeomorphic to a ball, we can then guarantee that $\omega_i=df_i$ for $f_i$ real functions on $U$.$^{(1)}$
Since the $\omega_i(p)$ are linearly independent, $(f_1,\cdots,f_n)$ is a chart around $p$. 
Given a tangent vector $X=\sum X_i \partial_i$, we know that $X_i=df_i(X)$. Therefore, it follows that $X(p)=\partial_1(p)$.

This is a slight overkill, we don't need Poincaré lemma. Since $\omega$ is constant, $\omega_i = \sum c_jdx_j$ for some constants $c_1,\cdots, c_n$. Letting $f_i= \sum c_jx_j$ yields an $f_i$ such that $df_i=\omega_i$.
A: I found the operative answer to my question (which show how to create this chart) 
If $|I|\neq1:$
$\psi_1=\eta_1\chi_{I^C}(1)+\sum_{i\in I}\frac{\eta_i}{|I|X_i(p)}$,
$\forall$ $1$<$i\leq n$, $\hspace{0.2cm}\psi_i=\begin{cases}
\sum_{j\in I-\{i\}}\frac{\eta_j}{X_j(p)}-\frac{(|I|-1)\eta_i}{X_i(p)},\hspace{0.2cm} X_i(p)\neq0\\
\eta_i,\hspace{0.2cm}X_i(p)=0
\end{cases}.$ 
Let $\mathcal{J}ac_p(\psi\cdot\eta^{-1})=\{a_{ij}\}_{i,j=1,...,n}.$
$a_{1,j}=\begin{cases}
\chi_{I^C}(1)+\frac{1}{X_1(p)}\cdot\chi_I(1)\hspace{0.2cm}j=1\\
\frac{1}{X_j(p)}\hspace{0.2cm}j\in I\hspace{0.2cm}j\geq2\\
0\hspace{0.2cm} otherwise
\end{cases}$ 
so from the formula above (in the question) $Y_1(p)=1$.
if $i$>$1$ fixed:
$a_{ij}=\delta_{ij}$ $\hspace{0.2cm}if\hspace{0.2cm} X_i(p)=0\Rightarrow$ $\hspace{0.2cm}Y_i(p)=X_i(p)=0$; 
alternatively:
$a_{ij}=\begin{cases}
-\frac{|I|-1}{X_i(p)} \hspace{0.2cm}if \hspace{0.2cm}i=j\\
\frac{1}{X_i(p)}\hspace{0.2cm} if \hspace{0.2cm}i\neq j
\end{cases}$ 
so $Y_i(p)=\sum_{j\in I-\{i\}}\frac{X_j(p)}{X_j(p)}-\frac{(|I|-1)X_i(p)}{X_i(p)}=|I|-1-(|I|-1)=0.$ Now we could easily check that the new component chart are linear combination of diffeomorphisms, so $\psi$ is a diffeomorphism. 
If $|I|=1\Rightarrow\exists\hspace{0.2cm}j\hspace{0.2cm}such\hspace{0.2cm}that\hspace{0.2cm}X_j(p)\neq0$ 
we put $\psi_1=\frac{\eta_j}{X_j(p)}$,
$\psi_i=\eta_{i-1}\hspace{0.2cm}if \hspace{0.2cm}2\leq i\leq j\hspace{0.2cm}$ 
$\psi_i=\eta_{i}\hspace{0.2cm}if \hspace{0.2cm}j+1\leq i\leq n$.
Since all the $X_i(p)$ are vanish except $X_j(p)$ and $a_{ij}\neq 0$ if and only if $i= 1$, so $Y_1(p)=1$ and $Y_i(p)=0\forall i\neq1.$
We could easily check that this jacobian determinant never vanish in $p$ and so this is a local diffeoporphism around $p$
