How many faces does the $n$-dimensional cube $I^n$ have? I am taking an online Coursera Calculus course, and this question popped up as one of the challenge problems, reproduced below.  I had a difficult time understanding the answer, which was $I^n$ has $2n$ faces.

This problem concerns the boundary operator $\partial$ from the bonus material.
  Denote by I the closed unit interval $[0,1]$. Then, as observed,
  $\partial I=\{0\}\cup\{1\}$ is the union of two points. Let's get a little creative.
  Denote by $I^n$ the "$n$-cube", that is, the Cartesian product of $n$
  intervals:
$$I^n = I \times I \times \cdots \times I$$
This is a well-defined and perfectly reasonable $n$-dimensional cube.
  (Just because you can't visualize doesn't mean it can't exist!) Note
  that $I^1=I$ and $I^0$ is a single point (a zero-dimensional cube!). As a
  step towards building a "calculus of spaces", let us write $∂I^1=2I^0=2$
  as a way of saying that the boundary of an interval consists of two
  points and that $I^0=1$.
The boundary of an n-dimensional cube consists of a certain number of
  $(n−1)$-dimensional cubes (called "faces"). For example, a square ($I^2$)
  has four faces. Using what you know about derivatives, answer this:
  how many faces does $I^n$ have?

 A: Hint: As you build the $n$ dimensional cube from the $n-1$ dimensional one, you extend it in a new direction.  As such, all the $n-2$ dimensional boundaries of the $n-1$ dimensional cube become $n-1$ dimensional boundaries of the $n$ dimensional cube.  You should be able to imagine this for $n=1,2,3$.  Use this to set up a recurrence relation that you can solve.
A: There are lots of ways to solve this problem.  The best is to find one for yourself.  If you don't have a clue, start with the cases which are easy to understand: a line segment has two vertices, a square has four sides, and a cube has six faces.  Now try to understand the connection between the dimension, $n$, and the number of faces, $2n$.
But here is one solution: a face is determined by choosing one coordinate $i$, with $1 \leq i \leq n$ and taking either all elements with $i$th coordinate $0$ or all elements with $i$th coordinate $1$.  (Thus for instance the top face of the unit cube $[0,1]^3$ is $\{ (x,y,1) \ | \ 0 \leq x,y \leq 1 \}$.)  So there are $2n$ faces altogether.
A: all comes from $\partial (A\times B) = \partial A B \cup A \partial B$
$\partial I^1 = 2$
To calculate the faces for $I^2$:
$\partial (I^2)=\partial (I^1 \times I^1) = \partial(I^1)\times I^1 \cup I^1\times \partial(I^1) = 2I^1 \cup 2 I^1$
therefore $2+2=4$ faces for $I^2$
To calculate the faces for $I^3$:
$\partial(I^1 \times I^2) = \partial I^1 \times I^2 \cup I^1\partial(I^2) = 2 \times I^2 \cup I^1 \times (2I^1 \cup 2 I^1) = 2I^2 \cup 2I^2 \cup 2I^2$
Therefore $I^3$ has $2+2+2$ faces.
following this reasoning, you can derive $I^n$
