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For a reference, you might see this paper of Birgit Richter. A rough outline follows:

Since $X = K(Z,n)$ can be made a commutative topological monoid per Ben Wieland's answer, its singular chain complex $C_*(X)$ is made into a commutative and associative differential graded algebra via the commutative associative Eilenberg-Zilber shuffle product $C_*(X) \otimes C_*(X) \to C_*(X \times X)$ composed with the multiplication on $X$.

The formula for the shuffle product is slightly involved when you iterate it, but essentially: to multiply $\alpha_1 \cdots \alpha_n$ with $\alpha_i$ of degree $k_i$, you sum over all ways to divide a set of size $\sum k_i$ into subsets of size $k_1, \cdots, k_n$ of a product of certain degeneracy operators, depending on the subdivision, applied to the chains $\alpha_i$. (With signs.)

If all the $\alpha_i$ are equal and in positive even degree $k$, then the signs don't interfere and we are summing over all ways to divide $nk$ into $n$ equal pieces of size $k$. However, because the chains are now equal we get the same term if we simply switch permute the pieces of size $k$, and so each term appears in the sum at least $n!$ times. (Note that $n k > 0$ is necessary here in order for these permutations to actually give different subivisions.) As a result, the n-fold product chain is divisible by $n!$.

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For a reference, you might see this paper of Birgit Richter. A rough outline follows:

Since $X = K(Z,n)$ can be made a commutative topological monoid per Ben Wieland's answer, its singular chain complex $C_*(X)$ is made into a commutative and associative differential graded algebra via the commutative associative Eilenberg-Zilber shuffle product $C_*(X) \otimes C_*(X) \to C_*(X \times X)$ composed with the multiplication on $X$.

The formula for the shuffle product is slightly involved when you iterate it, but essentially: to multiply $\alpha_1 \cdots \alpha_n$ with $\alpha_i$ of degree $k_i$, you sum over all ways to divide a set of size $\sum k_i$ into subsets of size $k_1, \cdots, k_n$ of a product of certain degeneracy operators, depending on the subdivision, applied to the chains $\alpha_i$. (With signs.)

If all the $\alpha_i$ are equal and in positive even degree $k$, then the signs don't interfere and we are summing over all ways to divide $nk$ into $n$ equal pieces of size $k$. However, because the chains are now equal we get the same term if we simply switch permute the pieces of size $k$, and so each term appears in the sum at least $n!$ times. (Note that $n > 0$ is necessary here in order for these permutations to actually give different subivisions.) As a result, the n-fold product chain is divisible by $n!$.