I am trying to show that the projection map $\pi: V_{k}(\mathbb{R}^{n+k}) \rightarrow \mathrm{Gr}_{k}(\mathbb{R}^{n+k})$ is a fiber bundle with fiber $O(k)$, the group of orthogonal $k \times k$-matrices. Here is what I have so far:
The Stiefel manifold $V_{k}(\mathbb{R}^{n+k})$ consists of all orthogonal $k$-tuples of vectors $(v_{1},...,v_{k})$ with $v_{i} \in \mathbb{R}^{n}$. We observe that the group $O(n)$ of orthogonal transformations acts transitively on the left of $V_{k}$. Then the isotropy at $(e_{1},...,e_{k})$ is $O(n-k)$ which is isomorphic to the $2 \times 2$ matrix with $I_{k}$ and $O(n-k)$ on the diagonal and zeroes elsewhere. This yields the identification $V_{k} \cong O(n)/O(n-k)$. We see that $V_{k}$ is a manifold from this, since its subspace topology coincides with that of $O(n)/O(n-k)$. Now we see that the group $O(k)$ acts on the right of $V_{k}$ via matrix multiplication. Under this action, two $k$-frames $v,v^{\prime}$ are in the same $O(k)$-orbit iff their vector space spans given by $\mathrm{span}(v) \subset \mathbb{R}^{n}$ and $\mathrm{span}(v^{\prime}) \subset \mathbb{R}^{n}$ are equal. Then, the orbit space $\mathrm{Gr}_{k}:=V_{k}=V_{k}/O(k)=O(n)/O(k) \times O(n-k)$ is our smooth Grassmanian manifold. Finally, the projection map $\pi$ that sends each $k$-frame $v$ to $\mathrm{span}(v)$ is the projection of a principal $O(k)$-bundle.
Is my understanding of this correct? Need I show anything further here? Any help would be appreciated. Thanks very much.
Here is a general fact that will yield the result you seek. If $G$ is a Lie group and $X$ is a manifold on which $G$ acts freely and properly, then $X/G$ has a natural manifold structure and the projection $X\rightarrow X/G$ is a principal $G$-bundle. In particular, $X\rightarrow X/G$ is a fibre bundle with fibre $G$.
Since $O(k)$ is compact, it automatically acts properly on $V_k$. You need only check that $O(k)$ acts freely on $V_k$, and this is indeed the case.
Added: Note that if $g\in O(k)$ stabilizes $(v_1,\ldots,v_k)$, then $g$ stabilizes $h(v_1,\ldots,v_k)$ for all $h\in O(n)$. In particular, $g$ stabilizes $(e_1,\ldots,e_k)$. Therefore, $g=I_k$. Hence, the $O(k)$-stabilizer of every $(v_1,\ldots,v_k)$ is trivial, meaning that the action is free.