tensor product of exterior algebras isomorphic to another exterior algebra

Question

In class (also on page 355 of Fomenko's Homotopical Topology), I learned that $$H^*(SU(n);\mathbb{Z})=\bigwedge[x_3,x_5,...,x_{2n-1}]$$ with $deg(x_i)=i$, and that $$H^*(SU(n);\mathbb{Z})\cong H^*(S^3\times S^5\times ...\times S^{2n-1};\mathbb{Z}).$$ However, I don't understand why that remark is true. For instance, if we set $n=3$, then we get the isomorphism $$\bigwedge[x_3,x_5]\cong H^*(S^3\times S^5)=\bigwedge[x_3]\otimes\bigwedge[x_5],$$ with the last equality due to Kunneth formula. It seems that these 2 rings have different multiplication rules if I send $(a_1+a_2x_3)\otimes (b_1+b_2x_5)$ to $a_1b_1+a_2b_1x_3+a_1b_2x_5+a_2b_2x_3x_5$. In the first ring, $x_3x_5=-x_5x_3$ while in the second ring $$(a_1\otimes a_2x_5)\cdot(b_1x_3\otimes b_2)=(b_1x_3\otimes b_2)\cdot (a_1\otimes a_2x_5).$$ I think I must have messed up the isomorphism between $\bigwedge[x_3,x_5]$ and $\bigwedge[x_3]\otimes \bigwedge[x_5]$, but I don't know what's the right isomorphism. Help?