Local Immersion Theorem in $\mathbb{R}^n$ proof I am trying to prove the following:

Let $U \subset \mathbb{R}^n$ be open and $f \in C^1(U;\mathbb{R}^m)$. 
Let $x^\star \in U, \ y^\star = f(x^\star)$. Suppose that
  $\mathrm{d}f(x^\star)$ is injective. Show that there are local
  coordinates around $x^\star$ and $y^\star$ such that $f(x_1, \dots,$
  $x_n) = (x_1, \dots, x_n, 0, \dots, 0)$ in those coordinates.

I don't know much about manifolds and I don't think that it is required for this result (since we are specifically in $\mathbb{R}^n$). I may apply Inverse / Implicit function theorems.
My definition of local coordinates about $x$ is a diffeomorphism $\xi \colon U_0 \rightarrow V_0$ between open neighbourhoods $U_0$ of $x$ and $V_0$ of $0$ in $\mathbb{R}^n$. (If you could explain what this means that would help greatly - I'm not sure how a diffeomorphism can be called coordinates).
The most useful answer would be some pointers for where to begin, so that I can do it myself. Thank you.
 A: I assume you want to show:
Given such $f=\begin{bmatrix}f_{1}\\f_{2}\\ \cdots\\f_{m}\end{bmatrix}$. There exist local coordinates
$$\xi_1:U_{1}\rightarrow V_{1}~with~x^{*}\in U_{1}\subset U,~0\in V_{1}~and~U_{1},V_{1}\subset \mathbb{R}^{n}$$
and
$$\xi_{2}:U_{2}\rightarrow V_{2}~with~f(x^{*})\in U_{2},~0\in V_{1}~and~U_{2},V_{2}\subset \mathbb{R}^{m}$$
such that
$$\xi_{2}(f(\xi_{1}^{-1}(x_1,x_2,\cdots,x_n)))=(x_1,x_2,\cdots,x_n,0,\cdots,0)$$
for all $(x_1,x_2,\cdots,x_n)\in V_{1}$.
Notice first we must have $n\leq m$ as part of our hypothesis.
The derivative of $f$ at $x$ is a $m\times n$ matrix 
$$Df(x)=\begin{bmatrix}D_{1}f_{1}&\cdots &D_{n}f_{1}\\D_{1}f_{2}&\cdots &D_{n}f_{2}\\ \cdots&\cdots&\cdots\\D_{1}f_{m}&\cdots &D_{n}f_{m}\end{bmatrix}.$$
It is injective as a linear map from $\mathbb{R}^{n}$ to $\mathbb{R}^{m}$. Hence there exists a $n\times n$ submatrix that is invertible. Without loss of generality we could assume this submatrix is 
$$\begin{bmatrix}D_{1}f_{1}&\cdots &D_{n}f_{1}\\D_{1}f_{2}&\cdots &D_{n}f_{2}\\ \cdots&\cdots&\cdots\\D_{1}f_{n}&\cdots &D_{n}f_{n}\end{bmatrix}.$$
Using inverse function theorem, it follows that $f_{0}=\begin{bmatrix}f_{1}\\f_{2}\\ \cdots\\f_{n}\end{bmatrix}$ is a local diffeomorphism around $x^{*}$.
Now define $\xi_{1}(x)=f_{0}(x)-f_{0}(x^{*})$ for $x$ around $x^{*}$. $\xi_{1}$ is a local coordinate in your definition. We have
$$\xi_{1}^{-1}(x_{1},\cdots,x_{n})=f_{0}^{-1}((x_{1},\cdots,x_{n})+f_{0}(x^{*}))$$
and
$$f(\xi_{1}^{-1}(x_{1},\cdots,x_{n}))=f(f_{0}^{-1}((x_{1},\cdots,x_{n})+f_{0}(x^{*})))=\begin{bmatrix}x_{1}+f_{1}(x^{*})\\x_{2}+f_{2}(x^{*})\\ \cdots\\x_{n}+f_{n}(x^{*})\\f_{n+1}(f_{0}^{-1}((x_{1},\cdots,x_{n})+f_{0}(x^{*})))\\ \cdots\\f_{m}(f_{0}^{-1}((x_{1},\cdots,x_{n})+f_{0}(x^{*})))\end{bmatrix}$$
Finally we define $$\xi_{2}(y_{1},\cdots,y_{n}\cdots,y_{m})=\begin{bmatrix}y_{1}-f_{1}(x^{*})\\y_{2}-f_{2}(x^{*})\\ \cdots\\y_{n}-f_{n}(x^{*}) \\y_{n+1}-f_{n+1}(f_{0}^{-1}((y_{1}-f_{1}(x^{*}),\cdots,y_{n}-f_{n}(x^{*}))+f_{0}(x^{*}))) \\ \cdots \\ y_{m}-f_{m}(f_{0}^{-1}((y_{1}-f_{1}(x^{*}),\cdots,y_{n}-f_{n}(x^{*}))+f_{0}(x^{*})))\end{bmatrix}$$
Please check $\xi_{2}$ is a local diffeomorphism and $$\xi_{2}(f(\xi_{1}^{-1}(x_1,x_2,\cdots,x_n)))=(x_1,x_2,\cdots,x_n,0,\cdots,0)$$
for all $(x_1,x_2,\cdots,x_n)\in V_{1}$ yourself by applying inverse function theorem, i.e looking at the derivative matrix, again.
