The period map is the integration of differential forms along cycles. More formally, we have the algebraic de Rham cohomology $H^*_{dR}(X)$ which is a Q-vector space. The Grothendieck--de Rham theorem proves that $H^*_{dR}(X) \otimes \mathbb C \simeq H_B^*(X) \otimes \mathbb C$, where $H_B^*(X)$ is the Betti cohomology, i.e. the rational singular cohomology of the topological space $X(\mathbb C)$. Over fields cohomology is dual to homology, we get a canonical nondegenerate pairing $H_{dR}^*(X) \otimes H^B_*(X) \to \mathbb C$ - the period map. Its image generates the ring of periods of $X$.

In this case there's something wrong with the statement of theorem M. Let $X = S^2$, then for all $k \gg 1$, $\pi^k(S^2) = 0$ while the corollary is false.

What do you mean by "of finite type"?

More like the last 2 pages imho. As I understand the argument: we consider holomorphic functions arising as solutions of algebraic regular differential equations associated with a flat connection. For a smooth proj. family of varieties the Gauss-Manin connection on $H^*_{dR}$ identifies the periods of fibers with a multi-valued solution of a diff. eq. on the base, the periods are values of these functions at rational points. The motivic Galois group associates nontrivial multivalued functions to any period. If one could show that $e$ isn't a value of such functions, we'd have a contradiction.

In principle it can be computed and gives an explicit complicated formula via all $\mathbf k$'s, but in fact it involves calculating all faces, edges, vertices etc, so it's more complex than the initial problem. It doesn't look like one could evaluate an integral of n-dimensional delta functions without computing intersections, but of course it's not a proof that some algorithm couldn't exist.

It's certainly possible to calculate the volume via Fourier analysis: $Vol(A)=\int 1_A dx$, where $1_A$ is the indicator function of $A$ equal to the product of $\Theta(\mathbf k \mathbf x -1)$ over all vectors $\mathbf k$ defining the linear inequalities. Here $\Theta$ is Heaviside's function. Since $\int 1_A dx = (\int 1_A e^{i p x} dx)|_{p=0}$ and the Fourier transform of a product is a convolution of Fourier transforms, we get a Fourier calculation for volume. The F.t. of each $\Theta$ is a sum of delta-functions on the space and on orthogonal to $\mathbf k$'s.

Finally, I describe a general approach to such algebraic questions. It may be an overkill for Hurewicz, but it is more powerful. E.g. I show that it proves Freudenthal, there are probably other interesting algebras which would be amenable to the method.

It's not a matter of innate complexity but a matter of historical precedence and inperfect axiomatization that $\infty$-categorical constructions are regarded as something vastly more difficult than classical ones. I also don't need that the universal $E_\infty$-group is THE sphere spectrum or BPQ theorem, it is mentioned only for comparison with other answer. The fact that groups can be delooped is a fundamental property of $\infty$-topoi, there is no getting around that one. Jeff also requires either that or Dold-Kan to construct $K(A, n)$. (...)

@DylanWilson Well, the OP has asked for an abstract nonsense proof. I have delivered on both points, didn't I? More seriously, the essential part of the proof is in the last paragraph. Most of the rest is required for the comparison to classic singular homology. The proof is trivial if you can define homology $\infty$-categorically in a simple way, e.g. if you know some simple description of the abelian operad (I don't). I consider Day convolution and operads to be a trivial matter. You're not terrified by Lawvere theories or coends, are you? (..cont)

@Oscar If you mean that I need Hurewicz or Freudenthal to prove that $\pi_0(\mathbb S) = \mathbb Z$, I don't. By the sphere spectrum I mean the universal $E_\infty$-group. It's 0-truncation is a universal abelian group, QED. If you wish, you can define $\mathbb S$ as the group completion of the groupoid of finite sets.

@OscarRandal-Williams No, because it's a single space of coefficients rather than homology itself. The real content of the proof is how you go from a single space to a colimit of spectra. Jeff Strom's answer relies on Freudenthal's theorem and stabilization, I proceed directly from the definition of free algebras. But of course all approaches will be somewhat related. I rely only on general category theory, thus I believe the proof works in any topos and for any algebraic theory, not only $\mathbb Z$-homology of spaces.

Hurewicz theorem is really the theorem about the homotopy category of spaces, thus the abstract $(\infty, 1)$-category of spaces is the proper setting for stating and proving it, and indeed it has a simple abstract proof. Is this what you're asking or are you really interested in the properties of the specific simplicial model? In the latter case you're bound to explicit computations.