Parallelizable spheres and whitney classes of bundles In Milnor's paper, He asserts the following corollary numbered 2

If $\mathbb{R}^n$ has bilinear product with no zero divisors, then $n=1,2,4$ or $8$.

In his proof he constructs a vector bundle $E$ on $S^n$ by clutching map $$x \mapsto (\text{left multiplication by }x),$$ and then asserts that that stiefel-whitney class $w_n(E) \neq 0$.
My question: How to prove that $w_n(E) \neq 0$ ?
 A: For definiteness, the definition of $E$ is that $E = (D^n\times \mathbb{R}^n)\cup_f (D^n\times \mathbb{R}^n)$ where $f$ is a map from $\partial(D^n\times \mathbb{R}^n) = S^{n-1}\times \mathbb{R}^n$ to the other copy given by $f(x,y) = (x,xy)$, where the product $xy$ is the bilinear product with no zero divisors.  The projection $\pi:E\rightarrow S^n$ maps the first $D^n$ to the north pole and the second $D^n$ so the south pole of $S^n$.
Because $GL_n$ deformation retracts onto $O(n)$, we may assume that for each $x\in S^{n-1}$, the map $y\mapsto xy$ is an isometry.  In particular, the multiplication map restricts to map $\mu:S^{n-1}\times S^{n-1}\rightarrow S^{n-1}$.
Proposition 1:  Under the obvious identification $H_{n-1}(S^{n-1}\times S^{n-1})\cong \mathbb{Z}\oplus \mathbb{Z}$, the induced map $\mu_\ast:H_{n-1}(S^{n-1}\times S^{n-1})\rightarrow H_{n-1}(S^{n-1})$ is, up to changing orientations, addition:  $\mu_{\ast}(a,b) = a+b$.
Proof:  Fix $y\in S^{n-1}$.  The self map of $S^{n-1}$ given by $x\mapsto xy$ is linear (since the multiplication is bilinear) and has trivial kernel (since there are no zero divisors), hence is an isomorphism.  In particular, this self map is a diffeomorphism so has degree $\pm 1$.  By fixing orientations correctly, it has degree $+1$.  As a consequence, we see the map $\mu_\ast(a,0) = a$.
Likewise, the self map given by $y\mapsto xy$ has degree $+1$, so $\mu_\ast(0,b) = b$.  Since $\mu$ is a homomorphism, we are done.  $\square$
Let $E'\subseteq E$ be the unit sphere bundle of $E$.  In terms of the description above, this simply means that we replace the copies of $\mathbb{R}^n$ with $S^{n-1}$.
Proposition 2:  $H_\ast(E')\cong H_\ast(S^{2n-1})$.
Proof:  We will use Mayer-Vietoris with $U = D^n\times S^{n-1}$, $V$ the other copy of $D^n\times S^{n-1}$.  Then $U\cap V$ has the homotopy type of $S^{n-1}\times S^{n-1}$.
Since $H_\ast(U) = H_{\ast}(V) = 0$ for $\ast \neq 0,n-1$, the Mayer-Vietoris sequence easily shows thats $H_\ast(E')\cong H_\ast(S^{2n-1})$ when $\ast \neq n-1, n$, so let's focus on those cases.
We get the exact sequence $$...\rightarrow H_n(U)\oplus H_n(V)\rightarrow H_n(E')\rightarrow H_{n-1}(U\cap V) \rightarrow H_{n-1}(U)\oplus H_{n-1}(V)\rightarrow H_{n-1}(E')\rightarrow H_{n-2}(U\cap V)\rightarrow ...$$ which is just $$0\rightarrow H_n(E')\rightarrow \mathbb{Z}\oplus \mathbb{Z}\rightarrow \mathbb{Z}\oplus \mathbb{Z}\rightarrow H_{n-1}(E')\rightarrow 0.$$
Let's figure out the map from $\mathbb{Z}^2$ to itself.  To that end, let $u\in H_{n-1}(U)$ and $v\in H_{n-1}(V)$ be generators, and let $a,b\in H_{n-1}(U\cap V)$ be generators corresponding to the two factors of $U\cap V \cong S^{n-1}\times S^{n-1}$.
I am going to view $U\cap V$ as living in $U$, and then attached to $V$ via $f$.  What this means this that the inclusion $U\cap V\rightarrow U$ induces the map $b\mapsto u$ and $a\mapsto 0$.
What about the map $U\cap V\rightarrow V$?  Well, this is really a composition $U\cap V\xrightarrow{f} U\cap V\subseteq V$.  From Proposition $1$, it easily follows that $f_\ast(a,b) = (a,a+b)$.  Since the inclusion $U\cap V\subseteq V$ maps $a$ to $0$ and $b$ to $v$, the composition is the map $a\mapsto v$ and $b \mapsto v$.
Returning to the Mayer-Vietoris sequence, the map $\mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}\times \mathbb{Z}$ is $a\mapsto (0,v)$ and $b\mapsto (u,v)$.  Now it is easy to verify that this map is a bijection, which finishes the proof of Proposition 2.  $\square$
Now, we'll compute the homology of $E'$ in a different way, and use this to compute $w_n(E)$.  To that end, recall the Grassmannian of $n$-planes in $\mathbb{R}^{\infty}$, $BO(n)$ has a tautological $n$-plane bundle over it, and that our bundle $E\rightarrow S^n$ is the pull back of this tautological bundle under some (homotopy class of) map $\phi:S^n\rightarrow BO(n)$.  Also, recall that $H^\ast(BO(n);\mathbb{Z}/2\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}[\omega_1,...,\omega_n]$ and that $w_i(E) = \phi^\ast(\omega_i)$.
Note that if we restrict the tautological bundle to its unit sphere and pull back, we get $E'$.
Proposition 3:  The unit sphere bundle in the tautological bundle over $BO(n)$ has total space homotopy equivalent to $BO(n-1)$.
Proof:  One definition of this tautological bundle is that we start with the $O(n)$ principal bundle $EO(n)\rightarrow BO(n)$ and use the Borel construction $\mathbb{R}^n\times_{O(n)} EO(n)$, where $O(n)$ acts on $\mathbb{R}^n$ in the usual fashion.  If we restrict this to the unit sphere bundle, we just get $S^{n-1}\times_{O(n)} EO(n)$.  Since the $O(n)$ action on $S^{n-1}$ is transitive with isotropy group $O(n-1)$, this space is homeomorphic to $EO(n)/O(n-1) = BO(n-1)$.  $\square$
So, now we have a commutative diagram of bundles $\require{AMScd}$
\begin{CD}
S^{n-1} @>>> S^{n-1}\\
@VVV @VVV\\
E' @>>> BO(n-1)\\
@VVV @VVV\\
S^n @>>> BO(n)\\
\end{CD}
Consider the $\mathbb{Z}/2\mathbb{Z}$-cohomology spectral sequence of the bundle on the right.  As shown in these notes, beginning on page 85, the differential $d_{n-1} = d$ has the property that $dx = \omega_n\in H^\ast(BO(n);\mathbb{Z}/2\mathbb{Z})$.  By naturality, it follows that in the bundle on the left, we have $d(x) = \phi^\ast(\omega_2) = w_2(E)$.
On the other hand, from Proposition 2 (and universal coefficients), we know that $H^\ast(E')\cong H^\ast(S^{2n-1})$.  Thus, in the spectral sequence for the bundle on the left, we must have $d(x)\neq 0\in H^n(S^n;\mathbb{Z}/2\mathbb{Z}) = E_{n-1}^{n,0}$.
Thus, $w_n(E) = d(x)\neq 0$, establishing Milnor's claim.
