A fiber bundle over Euclidean space is trivial.

Since you quote Bott--Tu, I'll use that:

They define on p.56 the pullback of a vector bundle. More generally you can define in the same manner the pullback of a fiber bundle $E$ on a space $B$ with fiber $F$ and structure group $G$: given a map $f : B' \to B$, there exists a new fiber bundle $f^{-1}E$ on $B'$ with again fiber $F$ and structure group $G$.

• Theorem 6.8 (p.57) extends to this setting: if $f \sim g$ ($f$ is homotope to $g$), then $f^{-1}E \cong g^{-1}E$ (the proof is essentially the same).

• It is also true that $(g \circ f)^{-1}E \cong f^{-1}(g^{-1}E)$, just like for vector bundles.

Now if $B$ is a Euclidean space, it's contractible, meaning there are map $f : B \to \{*\}$ and $g : \{*\} \to B$ such that $f \circ g \sim id_{\{*\}}$ and $g \circ f \sim id_B$. Now if you apply that last equality to any fiber bundle $E$ on $B$, you get $f^{-1}(g^{-1}E) \cong id^{-1}E = E$. But a fiber bundle on a singleton is necessarily trivial (directly from the definitions), thus $g^{-1}E$ is trivial. And the pullback of a trivial bundle is trivial again, so $E = f^{-1}(g^{-1}E)$ is trivial again.