Let $M$ be an $n$-dimensional, oriented, Riemannian manifold, $SO(TM)$ the oriented, orthonormal frame bundle of $M$ and $SM$ the sphere bundle of $M$. Let $\tilde{\pi}\colon SO(TM)\rightarrow SM$ (on the first tangent vector) and $\pi\colon SM\rightarrow M$ be the natural projections. There are canonical $1$-forms $\tilde{\theta}^i,\,i=1,...,n$ and $\tilde{\omega}_i^j,\,i,j=1,...,n$ on $SO(TM)$, characterized by $d\tilde{\theta}^i=\sum_{k=1}^n\tilde{\theta}^k\wedge\tilde{\omega}_k^i$ for $i=1,...,n$, $\tilde{\omega}_i^j=-\tilde{\omega}_j^i$ for $i,j=1,...,n$ and by the fact that, for any local section $s\colon U\rightarrow SO(TM)$ ($U\subseteq M$ open), $s^{\ast}\tilde{\theta}^1,...,s^{\ast}\tilde{\theta}^n$ form the coframe for the frame on $U$ given by the section. We have a natural connection $T(SO(TM))=V(SO(TM))\oplus H(SO(TM))$ given by $V(SO(TM))=\ker d(\pi\circ\tilde{\pi})$ and $H(SO(TM))=\ker\omega$, which pushes forward under $\tilde{\pi}$ to a connection $T(SM)=V(SM)\oplus H(SM)$ given by $V(SM)=\ker d\pi=d\tilde{\pi}(V(SO(TM)))$ and $H(SM)=d\tilde{\pi}(H(SO(TM)))$.
For $p\in M$, a choice of oriented, orthonormal basis of $T_pM$ yields a diffeomorphism $S_pM\cong S^{n-1}$, where $S_pM$ is the fiber of $SM$ at $p$. We can pull back the volume form on $S^{n-1}$ under this diffeomorphism to obtain a volume form on $S_pM$, which is independent of the chosen basis since two oriented, orthonormal basis differ by an action of $SO(n)$, under which the volume form on $S^{n-1}$ is invariant. For $v\in S_pM$, we can identify the tangent space $T_vS_pM$ with $V_{(p,v)}SM$ and extend the form by acting as $0$ on $H_{(p,v)}SM$, so that we ultimately obtain an $n-1$-form $\sigma$ on $SM$. This is the "fiber volume form".
Question: Why is $\tilde{\pi}^{\ast}\sigma=\tilde{\omega}_1^2\wedge...\wedge\tilde{\omega}_1^n$?