Questions in Hatcher, Algebraic Topology, Proposition 3B.1 http://pi.math.cornell.edu/~hatcher/AT/AT.pdf : This is the link to the book.
Proposition 3B.1. The boundary map in the cellular chain complex $C_*(X \times Y)$ is determined by the boundary maps in the cellular chain complexes $C_*(X)$ and $C_*(Y)$ via the formula 
$$d(e^i\times e^j)=de^i \times e^j+(-1)^i e^i \times de^j$$
I got stuck in the first paragraph of the proof:
Proof: Let us consider the special case of $I^n$. We give $I=[0,1]$ the CW structure with two vertices and one edge, so the $i^{th}$ copy of $I$ has a 1-cell $e_i$ and 0-cells $0_i$ and $1_i$ with $de_i=1_i-0_i$. The $n$-cell in the product $I^n$ is $e_1\times ...\times e_n$ and we claim that the boundary of this cell is given by the formula
$$d(e_1\times ...\times e_n)=\sum_i (-1)^{i+1}e_1\times ...de_i\times ...\times e_n ~~~~~(*)$$
$(*)$ is correct modulo the signs of the individual terms $e_1 \times ... 0_i \times ...\times e_n$ and $e_1 \times ... 1_i \times ...\times e_n$, since these are exactly the $(n-1)$-cells in the boundary sphere $\partial I^n$ of $I^n$.  To obtain the signs in $(*)$, note that switching the two ends of an $I$ factor of $I^n$ produces a reflection of $\partial I^n$, as does a transposition of two adjacent $I$ factors. Since reflections have degree $-1$, this implies that $(*)$ is correct up to an overall sign.
The boldface sentences are my questions (there are three of them).


*

*Must $\partial I^n$ coincide with the boundary of the $n$-cell in the CW sturcture of $I^n$?

*I see this only in intuition. How do I have to prove this?

*How does this hold?
Thanks in advance.
 A: (1) In this proposition Hatcher is doing an explicit computation for one CW complex, namely $I^n$ with the CW complex structure he describes. Recall that the CW structure for the product $A\times B$ has cells equal to products of cells in $A$ with those in $B$ with dimension equal to the sum of the dimensions. If you write it out carefully, you see that $I^n$ in this CW complex structure has one $n$-cell, which corresponds exactly to the interior of $I^n$ viewed as an $n$-cube, (the $n$-fold product of the interior of the interval with itself) and that its boundary is exactly $\partial I^n$.
(2) For definiteness, think of $I^n$ as the unit cube $[-\frac 12,\frac 12]^n$ in $\mathbb{R}^n$. Switching the two ends of one $I$ factor, say the $i$th, may be accomplished in $\mathbb{R}^n$ by the linear transformation 
$$ \begin{cases} e_i \mapsto -e_i \\ e_j \mapsto e_j & i\neq j. \end{cases} $$
Where  $e_1,\dotsc, e_n$ is the standard orthonormal basis  for $\mathbb{R}^n$. This linear transformation is a reflection about the coordinate hyperplane $x_i = 0$. Likewise, transposing two copies of $I$, say the $i$th and $(i+1)$st may be accomplished in $\mathbb{R}^n$ by the linear transformation
$$ \begin{cases}  e_i \mapsto e_{i+1}  \\ e_{i+1} \mapsto e_i \\ e_j \mapsto e_j & j \notin\{i,i+1\}.\end{cases}$$
This linear transformation is a reflection about the "diagonal" hyperplane $x_i = x_{i+1}$. Try writing this out for $n=2,3$ and see if that helps.
(3) The point of the preceding argument is the following: both kinds of linear transformation above send $\partial  I^n$ to itself. So assume that the sign of one of the $(n-1)$-cells appearing in the terms of the sum ($\ast$) is correct. By a sequence of reflections, we can send any other $(n-1)$-cell to that distinguished $(n-1)$-cell. By keeping track of the number of reflections required, we see that all of the signs appearing in ($\ast$) are correct if one of them is. If one of them is wrong, then the formula as a whole is off by a sign. (What follows is Hatcher's justification for why in fact the signs are correct.)
