If $f:A\rightarrow\Bbb{R}^n$ is a bijection such that $f\in C^r$ and $\det[Df(x)]\neq 0$ then $f[A]$ is open and $f^{-1}\in C^r$ What shown below is a reference from "Analysis on manifolds" by James R. Munkres.

So I don't understand why the function $\phi$ defined in the step $2$ is of class $C^r$: clearly $f(x)-c$ is of class $C^r$ but I don't know if the norm $||\cdot||$ and the square are of class $C^r$. Then I don't understand the final step $5$: indeed I don't understand why $I\in C^\infty$ and $Df\in C^{(r-1)}$. Indeed as reference I point out that Munkres said that if $f$ is a function mapping an open set $A$ of $\Bbb{R}^m$ into $\Bbb{R}^n$ then we say that $f$ is of class $C^r$ if and only if the partial derivatives of the functions $f_i$ of order less than or equal to $r$ are continuous on $A$, but unfortunately the function $Df:A\rightarrow GL(n)$ and $I:GL(n)\rightarrow GL(n)$ are not vector fields so I don't understand what Munkres says in step $5$. So to prove the statement I am only sure that I have to prove that the entries of $Dg$ are of class $C^{r-1}$. However perhaps I can explain the step $5$ as below. 
So could someone help me, please?
 A: First of all we know that the sum and and product of $C^r$ functions is a $C^r$ function so by permutation formula for determinant we know that for any square matrix $A$ it is 
$$
detA:=\sum_{\sigma\mathfrak{S}}sgn(\sigma)a_{1,\sigma(1)}\cdot...\cdot a_{n,\sigma(n)}
$$
and so if the elements of a square matrix are functions of class $C^r$ then the determinant is even a $C^r$ function; moreover by Cramer formula we know that if $B:=A^{-1}$ then 
$$
b_{i,j}=\frac{(-1)^{i+j}|A_{i,j}|}{|A|}
$$ 
and so if the elements of $A$ are $C^{r}$ functions then even the elements of $B$ are $C^r$ functions. 
Finally we remember that the composition of $C^r$ functions is even $C^r$. 
So we know that $f, g\in C^0$ and so $f\circ g\in C^0$ and so for what we observe the elements of $Dg(y)\equiv[Df(g(y))]^{-1}$ are continuous. So we assume that the theorem holds for functions of class $C^{r-1}$. Let $f$ be of class $C^r$ so that in particular $f$ is of class $C^{r-1}$ and thus (by the induction of hypothesis) the inverse function $g$ is of class $C^{r-1}$ and so even $f\circ g$ is of clas $C^r$; moreover the elements of $Df$ are functions of class $C^{r-1}$. Finally if $Dg(y)\equiv[Df(g(y))]^{-1}$ then for what we previously observed we conclude that the elements of $Dg$ are of class $C^{r-1}$ and so $g$ is of class $C^r$.
Well we remember that above we have said that the sum and product of $C^r$ functions is even $C^r$; moreover we can immediately observe that for the universal mapping theorem for products it follows that a vector field $f$ that mapping a subset $A$ of $\Bbb{R}^m$ into $\Bbb{R}^n$ is $C^r$ if and only if $f_i$ is $C^r$ for any $i:1,...,n$.
So we finally observe that if $f(x)$ is $C^r$ then clearly $\big(f_i(x)−c_i\big)$ is even $C^r$ so that we immediately conclude that $||f(x)−c||^2≡∑_{i=1}^n(f_i(x)−c_i)^2$ is $C^r$.
