A proof of Cartan-Dieudonné's theorem

As an assignment, we were asked to prove a theorem by Cartan and Dieudonné in the following form (the Euclidean space $\mathbb{E}^n$ is meant to be the usual $\mathbb{R}^n$ endowed with the usual inner product):

Every isometry of the Euclidean space $\mathbb{E}^n$ is the product of (at most) $n$ reflections about hyperplanes.

I proved it in the following way and I'd like to know whether it's a valid proof or not. I'd also like to know about other proofs (I'm sure there are...) :

We use induction on $n$. Let's assume the proposition for $n-1$. Let $\phi: \mathbb{E}^n\rightarrow \mathbb{E}^n$ be an isometry $\neq id$, and $\{e_1,\ldots, e_n\}$ be the canonical basis of $\mathbb{E}^n$. Consider the reflection $\sigma$ about the hyperplane orthogonal to the vector $e_1-\phi(e_1)$, which we assume non-zero, without loss of generality since $\phi\neq id$.

Clearly $\sigma(e_1)=\phi(e_1)$, so $e_1$ is ok. Since $\{\phi(e_2),\ldots, \phi(e_n)\}$ and $\{\sigma(e_2),\ldots, \sigma(e_n)\}$ are orthormal bases of the subspace $\langle \sigma(e_1)\rangle ^{\perp}$, there is an isometry $\psi: \mathbb{E}^n\rightarrow \mathbb{E}^n$ such that $\psi(\sigma(e_1))=\sigma(e_1)$ and $\psi(\sigma(e_j))=\phi(e_j)$, ($j\neq 1$). Since $\psi$ induces an isometry on $\langle \sigma(e_1)\rangle ^{\perp}$, by the inductive hypothesis, it can be written as a product of reflections about subspaces of $\langle \sigma(e_1)\rangle ^{\perp}$ and dimension $n-2$. Each of these subspaces can be extended to an hyperplane of $\mathbb{E}^n$ so that $\sigma(e_1)$ belongs to that and hence remains unchanged under those reflections. $\square$

-
You should specify that you mean linear isometries and reflections about hyperplanes passing through the origin (otherwise it will be $n+1$ instead, I suppose). –  tomasz Jul 4 '12 at 1:42