Are you saying that the solution in my question includes all the details but yours do not?

As I said, it implicitly prove that $\mathrm d \sigma _r=r^{n-1}\,\mathrm d \sigma $. As far as you have this formula, all details are included in my answer. @Brain

Actually, I do not understand, in which part of an advanced calculus course should this formula have been studied, could you clarify, please?

What is this formula trying to say?

It more or less says (in an infinitesimal way) that the area of the $n-$ball of radius $R$ is $R^{n-1}$ times the area of the unit ball (i.e. $|\partial B_R(0)|=R^{n-1}|\partial B_1(0)|$). For exemple, if $n=2$, then $|\partial B_1(0)|=2\pi$ and indeed $2\pi R=|\partial B_R(0)|=R |\partial B_1(0)|$. For $n=3$, you can see that $|\partial B_1(0)|=\frac{4}{3}\pi$ and indeed, $|\partial B_R(0)|=\frac{4}{3}\pi R^2=R^2|\partial B_1(0)|$. However, the proof in dimension $n$ is a bit long (as your solution shows).

Is there any text book (reference) that contains this proof?

The solution of your exercise ;-)

I need a reference ;-)

In the solution I posted, It is very confusing to me, the definition of $\textbf{F}(x)$, how the author is sure that it is a function from $\mathbb R^n$ to $\mathbb R^n$? could you clarify this please?

For your last comment, sorry, you are completely right, I should have said the unit $n-$sphere (and not the unit $n-$ball)... which is indeed the boundary of the unit $n-$ball... same for the $n-$sphere of radius $R$ (and not the $n-$ball of radius $R$). For your other comment, notice that $g(x)\in\mathbb R$ and $f(x)\in \mathbb R^n$ for all $x\in\mathbb R^n$. Therefore $F(x):=g(x)f(x)\in \mathbb R^n$ for all $x\in\mathbb R^n$, and thus indeed, $F:\mathbb R^n\to \mathbb R^n$. Let me know if something is still unclear. @Brain

Is the $i^{th}$ index correct in line 3 in the solution I posted, like, why there is no $i^{th}$ component on $g(x)$?

you meant $g(x) \in \mathbb R^n$, right? not $g(x) \in \mathbb R.$ but the author is subtracting $R^2$ from $g(x)$ which is just a number, right?

$g(x)\in \mathbb R$ (not $\mathbb R^n$... btw what would be the meaning of $g(x)f(x)$ if $f(x)\in\mathbb R^n$ and $g(x)\in \mathbb R^n$ ?) Notice that $x\cdot x\in \mathbb R$... that's why $g(x)\in \mathbb R$. And that's the reason why there is no $i^{th}$ component on $g$. @Brain

I am not sure what is the meaning of $|B_R(0)|$ and why the integration $\int_{B_R(0)} dx = |B_R(0)|$ ..... could you explain this to me please?

$|B_R(0)|$ is the volume of $B_R(0)$. And by definition, $\int_A\,\mathrm d x=|A|$.@Brain

I do not see where is the area of the unit ball is used at all in the solution I gave above. Could you clarify which line please? otherwise the solution contains a typo I guess.

But the solution is not using Fubini's theorem as you did, right?

I do not understand what is $\sigma$ exactly, is it a rectangular cross section of the sphere?

I only know two kinds of Fubini theorem, the one in "Calculus Early Transcendentals" and the other one in measure theory in "Real Analysis" by Royden & Fitzpatrick ...... the statment of Fubini you are using I do not understand it actually

The last property you are using, is there a reference for its proof?

1) As written in my OP, $\sigma $ is an element of $\partial B_1(0)$, but it's also the unit normal to the $n-$sphere. Also, $\mathrm d \sigma $ is an infinitesimal area of $\partial B_1(0)$. 2) Fubini theorem says that $$\int_{\mathbb R^n\times \mathbb R^m} f=\int_{\mathbb R^n}\int_{\mathbb R^m} f(x,y)\,\mathrm d y\,\mathrm d x=\int_{\mathbb R^m}\int_{\mathbb R^n}f(x,y)\,\mathrm d x\,\mathrm d y,$$ so, I used the first equality. 3) For your last question, why don't you try to prove it your self ;-)

My question, for the last property is, this is sth else other than that the area of the n-sphere of radius $R$ is $R^{n-1}$ the area of the unit sphere. right?

So in Fubini, $\mathbb R^n = [0,R]$ and $\mathbb R^m = \partial B_r(0) $?

1) Yes it's something else. Namely, the volume of the $n-$ball of radius $R$ is $R^n$ times the volume of the unit $n-$ball.

Ok, I see, I also posted another question related to this question, if you are interested in answering it please take a look at it.