This is a question regarding the first part of Spivak's proof of Stokes' theorem. Let $\omega$ be a $(k-1)$-form on $[0,1]^k$. Then $\omega$ is the sum of $(k-1)$-forms of the type $$ fdx^1\wedge\dots\wedge\hat{dx^i}\wedge\dots\wedge dx^k, $$ and it suffices to prove the theorem for each of these. Now Spivak claims the following (for the case $j=i$): $$ \int_{[0,1]^{k-1}}{I^{k}_{(j,\alpha)}}^*(fdx^1\wedge\dots\wedge\hat{dx^i}\wedge\dots\wedge dx^k) =\int_{[0,1]^k}f(x^1,\dots,\alpha,\dots,x^k)dx^1\dots dx^k. $$ Now, I know that a couple of posts have already been made on this part of the proof, but those were not helpful to me, so I will proceed to clarify my own confusion.
${I^{k}_{(j,\alpha)}}^*(fdx^1\wedge\dots\wedge\hat{dx^i}\wedge\dots\wedge dx^k)=({I^{k}_{(j,\alpha)}}^*f)({I^{k}_{(j,\alpha)}}^*dx^1)\wedge...\wedge\hat{dx^i}\wedge...\wedge({I^{k}_{(j,\alpha)}}^*dx^k)$
${I^{k}_{(j,\alpha)}}^*f=f\circ{I^{k}_{(j,\alpha)}}=f(x^1,\dots,x^{j-1},\alpha,x^{j+1},\dots,x^k)$
So i can have $$ \int_{[0,1]^{k-1}}{I^{k}_{(j,\alpha)}}^*(fdx^1\wedge\dots\wedge\hat{dx^i}\wedge\dots\wedge dx^k)=\int_{[0,1]^{k-1}}f(x^1,\dots,x^{j-1},\alpha,x^{j+1},\dots,x^k)({I^{k}_{(j,\alpha)}}^*dx^1)\wedge...\wedge\hat{dx^i}\wedge...\wedge({I^{k}_{(j,\alpha)}}^*dx^k)$$
But how should I proceed further to get the equation? how do we go from $[0,1]^{k-1}$ to $[0,1]^k$?
All he's doing is throwing in the extra $dx^i$ so that all the integrals over the faces of the cube are instead $k$-fold integrals over the entire cube. Why can he do this? Because $\int_0^1 dx^i = 1$ and we're integrating $f(x)$ with $x^i$ set equal to $\alpha$ (so that the function we're integrating is independent of $x^i$), the equality you copied follows. For example, $\int_{[0,1]} f(x,1)\,dx = \int_{[0,1]^2} f(x,1)\,dx\,dy$.