I know, for example, the every isometry of $\mathbb{R}^3$ can be written as a composition of at most $4$ reflections (through planes that doesn't necessarily have the 0 vector in them).
I wish to prove to more general statement that says every isometry of $\mathbb{R}^n$ is of the form of a composition of at most $n+1$ reflections.
The proof for the cases $n=2,3$ that I saw are a bit technical and I don't think that rotating and reflecting in $\mathbb{R}^n$ is a good way to show this (though possible and probably a lot more technical...)
I need help with my tactic to prove this : I want to use induction, knowing this is true for $n=1$.
The idea is this: if I were to define an isometry of the required form that acts the same on: the $0$ vector,$e_{1},...,e_{n}$ then I would be done.
Now I want to use, somehow, the induction hypothesis, my thoughts are that the given isometry, $f$, takes $e_{1},...,e_{n}$ into $n$ points and from the induction hypothesis there is a composition of at most $n$ reflections that take the projection of $e_{1},...,e_{n}$ to $\mathbb{R}^{n-1}$ to the projection of $f(e_{1}),...,f(e_{n})$.
From here I'm a bit lost, I want to say that this build also takes the zero vector of $\mathbb{R}^{n-1}$ to the projection of $f(0)$ to $\mathbb{R}^{n-1}$ and that I can compose what I got with one more reflection in a way that won't ruin what I already did and also help take care of the last coordinate.
Any thoughts ?
Edit: The proof in the link in the accepted answer it correct, but it seems that the proof given in the other answer is what I am trying to do, I would appriciate if someone could go into more details (I commented to the answer what I don't understand in it).
Edit 2: I tried writing the proof and complete all the details, did I do this correctly ?
We will prove by induction that every isometry of $\mathbb{R}^n$ can be represented as a composition of at most $n+1$reflections.
Base case: We know that the claim is true for $n=1$(and $n=2$but I think that this will follow).
Step: Let $V:=\left\{ x\in\mathbb{R}^n|x_{n}=0\right\} $and let $f\in Iso(\mathbb{R}^n)$.
If $f(0)\neq0$ then there exist $\tau\in O(n)$ s.t $\tau\circ f(0)=0$ (explicitly: $\tau$ is the reflection around the plane consistent of all point in $\mathbb{R}^n$ with equal distans from both $0$ and $f(0)$).
We now assume $f(0)=0$ (otherwise we denote $g=\tau\circ f$ and continue with $g$ ).
Consider $f|_{V}$ : $f$ is an isometry of $\mathbb{R}^n$ hence $\forall x,y\in\mathbb{R}^n:d(x,y)=d(f(x),f(y))$, in particular this holds for very $x,y\in V$ .
Let $E$ be the projection $E:\mathbb{R}^n\to\mathbb{R}^{n-1}$. $E(V)=\mathbb{R}^{n-1}$and $E\circ f|_{V}$ is an isometry of $\mathbb{R}^{n-1}$ hence be the induction hypothesis it can be represented as $R=R_{1}\circ...\circ R_{n-1}$.
We now expand $R$ to $\mathbb{R}^n$ in this manner: denote the matrix representin $R$ as $R_{M}$ ($R_{M}\in M_{n-1}(\mathbb{R})$) then $R$ expanded to $\mathbb{R}^n$ is represented by $\begin{pmatrix}R & 0\\ 0 & 1 \end{pmatrix}$that is : on the first $n-1$coordinates we act the same as $R$ did and we keep the last coordinate (note that $|R_{M}|=1$ and that it is straightforward to check that $R_{M}$ is an orthogonal matrix, $R_{1},...,R_{n}$are expanded from $\mathbb{R}^{n-1}$to $R^{n}$ in the same manner).
Denote $\varphi=R^{-1}\circ f$ (here $R$ is the expended isometry to $\mathbb{R}^n$). if $\varphi(e_{n})=e_{n}$ then we are done, otherwise there exist a reflection $\alpha$ s.t $\alpha\circ\varphi(e_{n})=e_{n}$and s.t $\alpha|_{V}\equiv f|_{V}$. (explicitly: $\alpha$ is a rotation around the plane consistent of all points $v\in\mathbb{R}^n$ s.t $\langle v,e_{n}-f(e_{n})\rangle=0$).
$\varphi$ is at most $n-1$ reflections hence $\alpha\circ\varphi$ is at most $n$ reflections.
In te first case $(f(0)\neq0)$ we are doing the process with $\tau\circ f$ hence in this case $f$ is represented by at most $n+1$ reflections.
Any comment about the style of the proof is also welcomed, this is one of the times I am writing a proof in English.