I am not sure if this qualifies as a simple argument, but to me it is very nice. I did not get the reference in the comments above - so sorry if this is close to that. I dont have an exact reference since this is pieced together from several places.
Let $A$ be any abelian group. Let $X=X(A)$ be the simplicial set defined by:
$X_p = \{$colorings of the $n$ faces of the standard $p$ simplex by elements in $A$ such that when restricted to any $n+1$ face the sum with alternating sign of the colorings of the faces are $0\}$
The face and degeneracy maps are defined by restriction and pull-back (the latter introduces 0's when an $n$ face is degenerate.
with this $X_p$ is a single point when $p < n$ (a single empty coloring), $X_{n}=A$, $X_{n+1}$ is "relations", and $X_p$ for $p>n+1$ is given by its restrictions to the $n+1$ faces. I.e. it is $n+(1 or 2?)$ co-skeletal. This means that its geometric realization $\lvert X \rvert$ is a $K(A,n)$.
In fact this is a simplicial group by adding the colorings, and $X(A)\times X(A)$ is isomorphic as simplicial groups to $X(A\oplus A)$ by definition.
To simplify we now use $A=\mathbb{Z}$ and to understand this product on the $\mathbb{Z}$ span of the simplices, I will assume some familiarity with the Eilenberg-Zilber operator, which appears naturally in the product of simplicial sets. Using this we describe the power map
(1) $H_n(X(A),A)^{\otimes k} \to H_{kn}(X(A)\times\cdots\times X(A),A)=H_{kn}(X(A^{\oplus k}),A) \to H_{kn}(X(A),A)$
$H_n(A,A)$$H_n(X(A),A)$ is a single $A$ and it is generated by the $n$ simplex $\alpha$ colored by $1$, the tensor product $\alpha^{\otimes k}$ can when mapped to the middle term of (1) be written as the sum
$\sum_{\sigma \in S(k,n)}$ sgn$(\sigma) \beta_{\sigma}$
where $S(k,N)$ is the permutations on $\{1,\dots,kn\}$ which preserves the order of the $k$ sequences SEQ$\strut_i=\{in+1,\dots,(i+1)n\}, i=0,k-1$ (a generalized shuffle), and $\beta_{\sigma}$ is the associated product of degenerations defined by
(2) $\beta_{\sigma}=(\sigma_1^* \alpha)\times \cdots \times (\sigma_k^* \alpha)$,
where $\sigma_i$ is the order preserving surjective map from $\{0,\dots,kn\}$ to $\{0,\dots,n\}$ defined by increasing one precisely when a number is in the image of $\sigma$ restricted to SEQ$\strut_i$.
In the case of $n$ even it is clear that the sign of the permutations $\sigma$ which simply permutes the sequences SEQ$\strut_i$ (without intertwining them) has sign 1 and thus we may act on the $\sigma$'s in the sum by these with out changing the sign, this action, however, permutes the factors in (2), but when mapped to the last factor in (1), they are the same, so the image of the sum in (1) is divisible by $k!$.
To see injectivity of the product one can see that the power map is injective on rational homology by using a Hopf-algebra argument as in section 3.C of Hatchers book.