How many verticies, edges and faces (cells) does an nd hypercube have? So I guess a 1d hypercube is a line segment.  It has 2 verticies and 1 edge.  Not sure how many faces it has?
A 2d hypercube is a square.  It has 4 verticies and 4 edges.  Again not sure how many faces it has?
A 3d hypercube is a cube.  It has 8 verticies and 12 edges and 6 faces.
A 4d hypercube has 16 verticies and not sure how many edges or 3d faces.
We can define the faces of an nd hypercube as being (n-1)d hypercubes.  So the faces of a cube are squares.
An nd hypercube has $2^n$ verticies.  (If we imagine a vertex coordinate $v = (v_1, v_2, ..., v_n)$ then choose some subset and set them to $1$, set the rest to $-1$.  This describes one vertex.  There are $2^n$ subsets of $n$ items.)
How many edges does it have as a function of n?  How do you derive that?
How many $(n-1)$d faces does it have?  How do you derive that?
Update
For the faces I think we can say: pick a dimension $i$, partition the verticies into two sets of $2^{n-1}$ verticies.  One with $v_i = -1$ and the other with $v_i = 1$.  Each of these two sets describes a face.  There are $n$ ways to pick $i$.  So an nd hypercube has $2n$ faces.  So a line segment has 2 faces (which are points).  A square has 4 faces (which are lines).  A cube has 6 faces (which are squares).  A 4d hypercube has 8 faces (which are cubes). and so on.
So just edges remains.
 A: Here are a couple hints. I will consider the vertices to be length-$n$ bitstrings.
An edge connects two vertices if the vertices differ by a single bit flip. How many edges are incident to a given vertex? If you multiply this by the number of vertices, what would you do next to end up with the total number of edges?
A facet (i.e. $(n-1)$-dimensional cell) may be chosen by first picking a coordinate, fixing its bit, and then letting all the other bits be arbitrary. How many ways are there to do this?
A: A hint: You can think of the $n+1$-hypercube as a $n$-hypercube, replicate it and move it a distance 1 orthogonal to the original hypercube. You can find a beautiful visualization here. Count the number of edges of the $n+1$ hypercube as the number of edges of the two $n$ hypercubes plus the number of edges needed to connect the two hypercubes. That will give you a recursive formula of the $k$-edge (there should be a more technical name) for the $n$ hypercube based on the number of $k$ and $k-1$ edges of the $n-1$ hypercube.
