Fiber bundle on Stiefel manifold Let $V_{n}(\mathbb{C}^{k})$ the Stiefel manifold of $n$-frame in $\mathbb{C}^{k}$. We can see $V_{n}(\mathbb{C}^{k})$ as a subset of $n$ copies of the cartesian product $S^{2k-1} \times \cdots \times S^{2K-1}$. So we have a bundle
$ V_{n-1}(\mathbb{C}^{k-1}) \rightarrow V_{n}(\mathbb{C}^{k}) \rightarrow S^{2k-1} $.
How can I find local trivializations?
 A: As I mentioned in the comments, given compact groups $H\subseteq K\subseteq G$, there is always a fiber bundle of the form $$K/H\rightarrow G/H \rightarrow G/K$$ where the projection map $\pi$ maps $gH$ to $gK$ and which has structure group $K$.
I don't know of any particularly easy or obvious proof of this fact.  The only reference I have for it is a bunch of hand written notes of my advisor, but there are probably other references out there.  The key lemma is that one always has a slice and that the isotropy representation in such a case must be trivial.
The most general form of it I know is: if $K$ is a compact Lie group acting smoothly on a manifold $M$ with all isotropy groups conjugate (say, to a subgroup $H$), then the projection $M\rightarrow M/K$ is a fiber bundle map with structure group $K$ and fiber $K/H$.
Anyway, using this on the inclusions $U(k-n)\subseteq U(k-1)\subseteq U(k)$ (via the usual block embeddings) gives you the fibration you're looking for.
For your particular example, if $G$ acts transitively on a manifold $M$ with isotropy subgroup $H$, then there is a $G$-equivariant diffeomorphism $G/H\cong M$. This is proven in Lee's Introduction to Smooth Manifolds Theorem 9.24.
So, in order to show, for example, $U(k)/U(k-1)\cong S^{2k-1}$, it suffices to find a transitive $U(k)$-action on $S^{2k-1}$ with isotropy subgroup $U(k-1)$.  Well, $U(k)$ naturally acts on $\mathbb{C}^k$ and preserves lengths of vectors, so it also acts on $S^{2k-1}\subseteq\mathbb{C}^k$. It's not too hard to see this action is transitive. The set of elements in $U(k)$ which fix the North pole is easily seen to be $U(k-1)$ block embedded.
For the Stiefel manfolds, we have $U(k)/U(k-n)\cong V_n(\mathbb{C}^k)$. Note that $U(k)$ naturally acts on $\mathbb{C}^k$, and hence on the collections of ordered tuples of vectors in $\mathbb{C}^k$. Given any $n$ orthonormal vectors $\{v_1,..., v_n\}$, complete them to an orthonormal basis $\{v_1,..., v_n, w_{n+1}, ... ,w_k\}$. The matrix whose columns are the $v$'s and $w$'s is an element of $U(k)$ which maps the first $n$ standard basis vectors to $\{v_1,...,v_n\}$.
This proves that $U(k)$ acts transitively on $V_n(\mathbb{C}^k)$. I'll leave it to you to check that the matrices in $U(k)$ which fix the first $n$ standard basis vectors are precisely the matrices of the form $\operatorname{diag}(I,B)$ with $B\in U(k-n)$.
