Why does the coefficients of the expansion of (2x+1)^n produce the elements of the hypercube? Why does the coefficients of the  expansion of $(2x+1)^n$ produce the elements of the hypercube?
For elements I mean the number of vertices, edges, square faces, cubes, hypercubes, etc that the next hypercube will have. For example for the tesseract:
$(2x+1)^4=16 x^4+32 x^3+24 x^2+8 x+1$
Meaning that the Tesseract has: 16 vertices, 32 edges, 24 square faces, 8 cubes and 1 tesseract as its elements.
 A: The $k$-th coefficient of $(2x+1)^n$ is $2^k\binom nk$. The $n$-cube can be modeled as the graph with vertices the elements of $\{0,1\}^n$ connected iff they differ by one digit. Using this, can you count the number of $k$-cubes an $n$-cube contains? Note it has $2^n$ vertices (the $0$-cubes) and $n2^{n-1}$ edges (the $1$-cubes). 

Spoiler The formula is $2^{n-k}\binom nk$. We can create a $k$-cube inside an $n$-cube as follows. First, pick a vertex. This can be done in $2^n$ ways. Now, pick $k$-spots in that vertex. Then start changing them. This gives a $k$-cube. But this $k$-cube has $2^k$ vertices, and any of those would have worked in obtaining it. Thus, the final result amounts to $2^n\binom nk/2^k=2^{n-k}\binom nk$.

A: There is no need to appeal to the binomial theorem; aim to see the result directly. The point is that we can think of an $n$-dimensional cube as the cartesian product $[0, 1]^n$ of $n$ copies of the $1$-dimensional cube, that we can break up $[0, 1]$ into two vertices $\{ 0 \}, \{ 1 \}$ and an open edge $(0, 1)$, and that cartesian product distributes over disjoint union. For example, when $n = 2$,
$$\left( \{ 0 \} + (0, 1) + \{ 1 \} \right)^2 = \{ 0 \} \times \{ 0 \} + \{ 0 \} \times (0, 1) + \{ 0 \} \times \{ 1 \} + (0, 1) \times \{ 0 \} + \dots$$
This is best explained by a picture but I can't find an existing picture that does what I want. In any case it's better to think of the result as an expansion of $(2 + e)^n$ where $e$ stands for the open edge so that the exponent of $e$ matches with the dimension of the corresponding face. 
A: The number of $m$-dimensional cubes on a $n$-dimensional cube is given by
$$
2^{n-m}\binom{n}{m}
$$
(see Wikipedia).
Expanding your polynomial (via the Binomial theorem) gives the same numbers as above for the coefficients.
