Prove that Steifel Manifold as a Manifold Using Group Action

Question

The Stiefel Manifold is the space of $k$-tuples of orthonormal vectors in $\mathbb{R}^{n+k}$ $$ V_k\left(\mathbb{R}^{n+k}\right)=\left\{\left(v_1, \ldots, v_k\right): e_i \in \mathbb{R}^{n+k},\left\langle v_i, v_j\right\rangle=\delta_{i, j}\right\} $$

I am trying to show that Steifel Manifold can be endowed with a smooth structure by relating it to $O(n+k)$ i.e. realizing $O(n)$ as a submatrix of $O(n+k)$

$$ O(n) \hookrightarrow O(n+k), \quad O(n) \ni B \mapsto\left(\begin{array}{cc} I_{k, k} & 0_{k, n} \\ 0_{n, k} & B \end{array}\right) $$ Then $O(n)$ group acts on $O(n+k)$ is: $$ \left(\begin{array}{ll} A_{k, k} & A_{k, n} \\ A_{n, k} & A_{n, n} \end{array}\right) \cdot\left(\begin{array}{cc} I_{k, k} & 0_{k, n} \\ 0_{n, k} & B \end{array}\right)=\left(\begin{array}{ll} A_{k, k} & A_{k, n} B \\ A_{n, k} & A_{n, n} B \end{array}\right) $$ The action is free (?) and proper. If I can prove $$ \pi: O(n+k) \rightarrow V_k\left(\mathbb{R}^{n+k}\right) $$ as a projection $O(n+ k) \xrightarrow{i} \mathbb{R}^{(n+k)^2}\to V_k\left(\mathbb{R}^{n+k}\right)\subset \mathbb{R}^{(n+k)\times k} $ that satisfies:

$$\pi(Q_1) = \pi(Q_2) \leftrightarrow Q_2 = Q_1 B \quad \text{for some B} \in O(n)\quad (*)$$ Then a bijection $$O(n+k)/O(n) \cong V_k\left(\mathbb{R}^{n+k}\right)$$ can be established. Upgrade this bijection to diffeomorphism is quite easy and it is not in the scope of this question.

A second thought raises 2 questions:

1. Why the action is free? In other words, $A_{k,n} B = A_{k,n} \Rightarrow B= I$. But is it the case here?
2. Denote the submatrix of $Q_1,Q_2$ as $(Q_1)_{k,n}$ and $(Q_2)_{k,n}$. Is it true that there exists $B \in O(n)$ such that $(Q_2)_{k,n} = (Q_1)_{k,n} B$ ?

Both of these 2 questions are just linear algebra. But I could not think out a property of orthogonal matrix that can be applied here.

Edit: Maybe I overthink this problem... I do not think these sub-matrix are actually relevant for this proof.

Thanks in Advance!

Answer: 1) Let $A \in O(n+k), B \in O(n)$ such that $AB = I$. Then since $I = (AB)^T(AB)$, $B = I$, which means the action is free.
2) Let $F: O(n+k)/O(n) \to V_k$ defined to be $\pi_F(Q O(n)) = F([Q])$. Surjectivity follows from the definition of $\pi$. The non-trivial part is to prove the injectivity i.e. to prove (*): If $\pi(Q) = \pi(M)$, $Q$ and $M$ agrees on the first k-columns. The remaining $n$ columns constitute a $n$-orthonormal tuples in $\mathbb{R}^{n+k}$. Denote $\alpha$ to be the last $n$-columns for $Q$ and $\beta$ to be the last $n$-columns for $M$. Then a linear transformation from an orthonormal basis to an orthonormal basis is an orthogonal transformation, with a corresponding matrix orthogonal basis $B \in O(n)$. Since the transformation does not change the first $k$ columns, we have $Q = M \cdot B$.