Prove that $f(W)$ is the graph of $y_{n+1} = \varphi(y_1,\cdots,y_n)$ 
Let $f: U \subseteq \mathbb R^n \rightarrow \mathbb R^{n+1}$ be of class $C^k,k\geq 1,$ and $U$ open. If for every $x\in U$,
$$f(x) = (f_1(x),\cdots, f_{n+1}(x)) \text{    and   }\det\bigg(\frac{\partial f_i}{\partial x_j}\bigg)_{1\leq i,j \leq n} \neq0,$$
then, for every $x\in U$, there exists a neighbourhood $W\subseteq U$ of $x$ such that $f(W)$ is the graph of a $C^k$ function $y_{n+1} = \varphi (y_1,\cdots,y_n)$.

Saying that $f(W)$ is the graph of $\varphi$ is the same as saying the following sets are equal:
$\{(f_1(x),\cdots,f_{n+1}(x)): x\in W\} = \{((y_1,\cdots,y_n,\varphi(y_1,\cdots,y_n)): (y_1,\cdots,y_n)\in \text{Domain of $\varphi$}\}$.
In other words, I have to prove that locally there is a function $\varphi$ depending on the previous coordinates $f_1,\cdots,f_n$.
The first thing that I've noticed is that $f'(x)$ is an injective linear transformation. Indeed, we have $n = \dim \ker f'(x) + \dim \text{im}f'(x) \geq n + \dim \ker f'(x) \geq n \implies \dim\ker f'(x) =0,$ since $f'(x)$ has at least $n$ linearly independent lines.
Now I don't know how to proceed exactly. Initially, I was wondering of using the local immersion theorem (since $f'(x)$ is injective), but I couldnt see a way to use this theorem to express $f_{n+1}$ in terms of the others.
I also considered the function $\pi:\mathbb R^{n+1} \rightarrow \mathbb R^n, (x_1,\cdots, x_{n+1}) \mapsto (x_1,\cdots, x_n).$
So, $\pi(f(x)) = (f_1(x),\cdots,f_n(x)).$ Writing $g = \pi \circ f$, its derivative $g'(x)$ is invertible, hence, it is a local diffeomorphism with inverse $h$. Therefore, $\pi \circ f \circ h = g \circ h = I_d$ and we have $\pi(f(h(x_1,\dots,x_n) ) = (x_1,\cdots,x_n).$ If I could "get rid" of $\pi$ somehow, this equation would give me that $f(h(x_1,\cdots,x_n)) = (x_1,\cdots,x_n,\varphi(x_1,\cdots,x_n))$ and thats what whe need to show. But I cant find a clear way to say or jusitfy this.
Any insight, hint? Thank you.
 A: You are very near the truth.
I shall denote the points $y\in{\mathbb R}^{n+1}$ by $(y',y_{n+1})$ with $y'=(y_1,\ldots, y_n)$, and let $\pi:\>{\mathbb R}^{n+1}\to{\mathbb R}^n$ be the projection forgetting the last coordinate.
Choose an arbitrary point $p\in U$, and let $f(p)=:q=(q',q_{n+1})$.  Since we have everywhere $${\rm det}\left({\partial f_i\over\partial x_k}\right)_{1\leq i,\,j\leq n}\ne0$$
there is a neighborhood $W$ of $p$ such that the map $$f':=\pi\circ f=(f_1,f_2,\ldots, f_n)$$ maps $W$ diffeomorphically onto a  neighborhood $V\subset{\mathbb R}^n$ of the point $q'\in{\mathbb R}^n$. There is a $C^1$-inverse $$g:=\bigl(f'\bigr)^{-1}:\quad V\to W\ .$$
The $C^1$ function
$$\phi:=f_{n+1}\circ g:\quad V\to{\mathbb R}$$
gives for each point $y'\in V$ the last coordinate $y_{n+1}$ of a point $y=(y',y_{n+1})\in{\mathbb R}^{n+1}$. The graph of this $\phi$ is the set
$${\cal G}=\bigl\{\bigl(y',\phi(y')\bigr)\in{\mathbb R}^n\times{\mathbb R}\bigm| y'\in V\bigr\}\ .$$
Note that $f(W)=(f\circ g)(V)$. From
$$(f\circ g)(y')=\bigl((f'\circ g)(y'),(f_{n+1}\circ g)(y')\bigr)=\bigl(y',\phi(y')\bigr)\qquad(y'\in V)$$
it finally  follows that indeed $f(W)={\cal G}$.
