In Spivak's Calculus on Manifolds, the proof of Stokes Theorem on $\mathbb{R}^n$ begins as follows...
It seems to me that there's something here which can be very confusing: When you pull back the $k-1$ form $f dx^1 \wedge ... \wedge \widehat{dx^i} \wedge ... \wedge dx^k$ along ${I^k}_{(i,\alpha)}$, the result is again a $k-1$ form, which should be integrated over a $(k-1)$-cube.
However, in the line below, the integral is over $[0,1]^k$. It amounts to the same thing, since ${I^k}_{(i,\alpha)}^*(f dx^1 \wedge ... \wedge \widehat{dx^i} \wedge ... \wedge dx^k) = f(x^1, ..., x^{i-1},\alpha,x^i, ..., x^{k-1})\,dx^1 \wedge ... \wedge dx^{k-1}$ and then
$$ \begin{aligned}& \int_{[0,1]^{k-1}}f(x^1, ..., x^{i-1},\alpha,x^i, ..., x^{k-1})\,dx^1 \wedge ... \wedge dx^{k-1} \\ = & \int_{[0,1]}\left(\int_{[0,1]^{k-1}}f(x^1, ..., x^{i-1},\alpha,x^i, ..., x^{k-1})\,dx^1 \wedge ... \wedge dx^{k-1}\right)dx^k \\ = & \int_{[0,1]^{k}}f(x^1, ..., x^{i-1},\alpha,x^i, ..., x^{k-1})\,dx^1 \wedge ... \wedge dx^k \\ = & \int_{[0,1]^{k}}f(x^1, ..., x^{i-1},\alpha,x^{i}, ..., x^{k-1})\,dx^1 ... dx^k \\ = & \int_{[0,1]^{k}}f(x^1, ..., x^{i-1},\alpha,x^{i+1}, ..., x^{k})\,dx^1 ... dx^k\end{aligned}$$
where the second line follows since the pulled back form is constant with respect to $x^k$ and the last line follows since we're working with the Riemann integral over $[0,1]^k$ so we're really just renaming variables. I think it's a bit of a stretch to ask the reader to 'note' that without any further indication as to why it's true.
Spivak pulls a similar trick later on in the proof, which I noticed another StackExchange question on. After having run through the steps of the proof on a small example, I'm guessing that the reason for doing this is to avoid having to talk about renaming variables.
So, my two questions are:
Is there a simpler way to make sense of the 'note' which I addressed above?
Am I correct in thinking that the extra integration is done to make the proof more concise and avoid discussion of renaming variables? Or is there some other reason I'm missing?
I agree that it's conceptually somewhat unsatisfying to turn the $(k-1)$-dimensional integrals into $k$-dimensional integrals, but it avoids all sorts of ugly notation. Note, for example, that in the second line of your second paragraph, you got it wrong: You should have written $$(I^k_{i,\alpha}){}^*f dx^1\wedge\dots\wedge\widehat{dx^i}\wedge\dots\wedge dx^k = f(x^1,\dots,x^{i-1},\alpha,x^{i+1},\dots,x^k)dx^1\wedge\dots\wedge\widehat{dx^i}\wedge\dots\wedge dx^k.$$ Rewriting all the integrals over the $k$-cube avoids the notational morass. (It's not a matter of "remaining" variables; it's a matter of notating which one is omitted. But most likely that's what you intended.)