n-cubes vs simplices when defining chains When defining chains, the standard definition is formal linear combinations of n-simplices. However in Calculus on Manifolds by Spivak, he defines chains as formal linear combinations of n-cubes.
I was trying to figure out why would you choose one definition over the other. As such, the definition of a singular n-cube is easier (to me anyway), as a continuous function from [0,1]^n to some space, whereas the definition of a k-simplex is sufficiently complicated. 
Why is there a difference, how does it affect the boundary map and does it affect how we integrate on chains?
 A: $n$-cubes have the advantage that integrating over them (which is what Spivak wants to do) is easy, and another advantage (that Spivak doesn't really use) which is that the "cartesian product" of an $n$-cube and a $k$-cube is an $n+k$ cube. 
On the other hand, simplices have their own advantages, like simplicity in various contexts. 
Because every simplex can be realized as an $n$-cube (for a triangle, consider the map $(s, t) \mapsto (st, t)$ that sends a unit square to the traingle with verts $(0,0), (1, 0), (0, 1)$), and every $n$-cube can be represented as a simplicial chain (for a square, draw a diagonal to represent it as a sum of two simplices, for instance), the resulting algebraic-topological structures (like homology groups) end up being the same. So: it's a matter of convenience for the thing you're doing at the time. Spivak's trying to get to a proof of Stokes' Theorem in about 12 pages, which is, I suspect, why he uses $n$-cubes. 
(It may also be tied to why he uses products of intervals in defining open sets rather than using disks, as is more traditional, but I haven't really checked that.) 
