Nice question. Here's Unfortunately the answer I posted an argument that hour ago is fairly elementary, assuming one knows wrong because I switched the structures of homology and cohomology ring structures of for the James reduced product$J(S^{2n})$. (An elementary exposition of these can be found in my algebraic topology book.) There is a map $f:J(S^{2n})\to K({\mathbb Z},2n)$ that is an isomorphism on $H_{2n}(-;{\mathbb Z})$, and this can be taken to be an H-map since $J(S^{2n})$ . It is the free monoid generated by $S^{2n}$ and $K({\mathbb Z},2n)$ is an H-space. Since $J(S^{2n})$ cohomology that is a divided polynomial ring on a generator in degree $2n$, it will suffice to show that $f$ is injective on integer homology. This is equivalent to $f$ being nonzero in rational homology in each degree $2nk$. In rational cohomology $f$ is surjective in all degrees since while the rational cohomology of $J(S^{2n})$ homology is a polynomial ringon a generator in degree $2n$ and $f$ is an isomorphism in that degree. Since $f$ is surjective in rational cohomology, it is nonzero in rational homology in degrees $2nk$, which finishes the argument.
Nice question. Here's an argument that is fairly elementary, assuming one knows the homology and cohomology ring structures of the James reduced product $J(S^{2n})$. (An elementary exposition of these can be found in my algebraic topology book.) There is a map $f:J(S^{2n})\to K({\mathbb Z},2n)$ that is an isomorphism on $H_{2n}(-;{\mathbb Z})$, and this can be taken to be an H-map since $J(S^{2n})$ is the free monoid generated by $S^{2n}$ and $K({\mathbb Z},2n)$ is an H-space. Since $J(S^{2n})$ is a divided polynomial ring on a generator in degree $2n$, it will suffice to show that $f$ is injective on integer homology. This is equivalent to $f$ being nonzero in rational homology in each degree $2nk$. In rational cohomology $f$ is surjective in all degrees since the rational cohomology of $J(S^{2n})$ is a polynomial ring on a generator in degree $2n$ and $f$ is an isomorphism in that degree. Since $f$ is surjective in rational cohomology, it is nonzero in rational homology in degrees $2nk$, which finishes the argument.