How many parameters are needed to specify a k-dimensional subspace of R^d? What is the number $N^d_k$ of real-valued parameters that are needed to specify a k-dimensional subspace of $\mathbb{R}^d$? And how can these parameters be interpreted?

I know: $N^d_1 = N^d_{n-1} = d - 1 = \binom{d}{1} - 1$.
The parameters can be interpreted as the d components of a vector spanning the 1-dimensional subspace minus its (arbitrary) length.
I know: $N_1^3 = N^3_2 = 2 = \binom{3}{2} - 1$. 
The parameters can be interpreted as two angles or as the three components of a normal vector of the 2-dimensional subspace minus its (arbitrary) length.
I know: $N^d_2 = N^d_{d-2} = \binom{d}{2} - 1$
I believe this, because a d-dimensional rotation has $\binom{d}{2}$ degrees of freedom, one for the rotation angle, the remaining $\binom{d}{2} -1$ ones for the (d-2)-dimensional (hyper)plane of rotation which also defines a 2-dimensional hyperplane as its orthogonal complement.
Question: How do I know that $\binom{d}{2}$ is the number of degrees of freedom of a d-dimensional rotation? 
How can these $\binom{d}{2}$ parameters of a rotation or the $\binom{d}{2} - 1$ parameters of a 2-dimensional hyperplane be interpreted (maybe even intuitively)?

I guess that $N^d_k$, the number of parameters that are needed to specify a k-dimensional subspace of $\mathbb{R}^d$, is given by $\binom{d}{k} -1$. How can this be shown? Only formally by mathematical induction or more directly, using e.g. the observation, that there are $\binom{d}{k}$ k-dimensional subspaces of $\mathbb{R}^d$ spanned by k of d elements of an orthonormal basis of $\mathbb{R}^d$?
 A: This answer is similar to what Ben said about $d \times k$-matrices, but maybe a little more visual.  Instead of computing the dimension "as a whole", I'll just compute the dimension of subspaces neighboring a given one.
Take some fixed $k$-dimensional subspace $P$ with complementary space $P^\perp$ of dimension $n-k$.  Any sufficiently nearby subspace $P'$ to $P$ looks like a graph of a linear function $A : P \to P^\perp$.  And on the other hand, any such linear function defines a unique subspace.  So you only need to count linear maps from $P$ to $P^\perp$, which are $k(n-k)$-dimensional.

Thinking about the issue locally also helps avoid a mistaken dimension count like $n \choose k$.  The problem with the $n \choose k$ count is that you are enumerating some random points on the Grassmannian, which doesn't tell you anything about the dimension.  For example, take the case of 2-dimensional subspaces of $\mathbb{R}^3$.  A subspace here is determined by its normal vector, so there is a bijection between 2-dimensional (oriented) subspaces and the unit sphere.
The $3 \choose 2$ planes in your count correspond to the north pole and two equatorial points $90^\circ$ apart.  But you wouldn't conclude that the sphere is 3-dimensional just because it has three points!
A: Let me expand a little on Ben's matrix remark, since all this talk about Grassmannians might give you the impression that the answer is complicated!  Any $d \times d$ matrix has a rowspace of dimension $k$, the row rank of the matrix.  You learned in linear algebra that


*

*row operations leave the rowspace of a matrix invariant, and

*up to row operations, every matrix has a unique representative in reduced row echelon form.


I claim that this allows us to identify $k$-dimensional subspaces of $\mathbb{R}^d$ with rrefs with $k$ nonzero rows.  Why?  Because any two bases of a given space, when written down in matrix form, are related by row operations.
Now the rrefs with $k$ nonzero rows are exactly as Ben describes them: the $k \times k$ box in the upper left is the identity matrix and the $k(d-k)$ entries to the right of those can be chosen freely.  (The fancy term for this is called the "Schubert cell decomposition" of the Grassmannian.)
A: The answer to this question is $k(d-k)$ "parameters" actually.  One way to make this formal is to say that the Grassmannian of k planes in $\mathbb{R}^d$  is a $k(d-k)$ dimensional manifold.  
One way to think about this result is that generically, such a plane is the image a unique $d\times k$-matrix where the first $k\times k$ box is the identity matrix, leaving $k(d-k)$ entries you can choose freely. 
You may want to look up some information on "Grassmannians" since there is much more detailed stuff out there on this question.
A: Here is a more geometric way to find the answer. We can specify a $k$ dimensional subspace by giving $k$ vectors in that subspace. That would require $kn$ parameters, but we are overspecifying. By how much are we overspecifying? Write those vectors in a basis such that the first $k$ coordinates specify the location in the subspace, and the remaining $n-k$ coordinates specify the location orthogonal to the subspace. For specifying a subspace the $k$ coordinates within the subspace do not matter, since whatever values for those coordinates we choose, we get the same subspace. Therefore only the $n-k$ components of those $k$ vectors matter, so that's a total of $k(n-k)$ parameters.
In my opinion such a geometric argument gives far more insight than looking at reduced row echelon form matrices.
