The signature is multiplicative under Cartesian products

Kunneth decomposition: by [F2], $H^{2k}(M\times N;\mathbb{R})={⨁}_{i=0}^{2k}{V}_i$ with $V_i=H^i(M;\mathbb{R})\otimes H^{2k-i}(N;\mathbb{R})$, and the ring structure is the graded tensor product.

For $x\otimes y\in V_i$ and $x\mathrm{\prime }\otimes y\mathrm{\prime }\in V_j$ with $i+j=4a$, the ring formula [F2] and the matching-degree evaluation [F3] give $Q_{M\times N}(x\otimes y,x\mathrm{\prime }\otimes y\mathrm{\prime })=(-1{)}^{(2k-i)j}⟨x⌣x\mathrm{\prime },[M]⟩⟨y⌣y\mathrm{\prime },[N]⟩$. If $i+j\mathrm{\ne }4a$, one factor cup class has degree greater than the dimension of its manifold, so vanishes by [F2]. Thus $V_i$ pairs only with $V_{4a-i}$; the factors in the displayed formula are complementary-degree cup pairings, rather than middle forms unless $i=j=2a$.

Middle block: taking $i=j=2a$ in step 1.2, the sign is $(-1{)}^{\mathrm{\mid }y\mathrm{\mid }\mathrm{\mid }{x}^{\mathrm{\prime }}\mathrm{\mid }}=(-1{)}^{2b\cdot 2a}=+1$, so $Q_{M\times N}$ restricted to $V_{2a}=H^{2a}(M)\otimes H^{2b}(N)$ is exactly $Q_M\otimes Q_N$.

Off-middle part is nondegenerate: by [F4] $\dim V_i=\dim V_{4a-i}$ for every $i$, and the pairing of $V_i$ with $V_{4a-i}$ induced by $Q_{M\times N}$ is the tensor product of the perfect pairings $H^i(M)\times H^{4a-i}(M)\to \mathbb{R}$ and $H^{2k-i}(N)\times H^{4b-2k+i}(N)\to \mathbb{R}$, hence perfect: in dual bases the tensor pairing matrix is a nonzero scalar times an identity matrix, the scalar being the fixed Koszul sign. Set $V_i=0$ when $i$ is outside $0,\dots ,2k$; if a block has no complementary index in that range it is zero by the dimension bounds in [F2]. Therefore $Z={⨁}_{i\mathrm{\ne }2a}{V}_i$ is nondegenerate: a class in $Z$ pairing to zero with all of $Z$ must have every block component zero, testing against the complementary block.

Half-dimensional isotropic subspace: $W={⨁}_{i<2a}{V}_i\subseteq Z$ is totally isotropic by step 1.2, since $i,j<2a$ give $i+j<4a$ and hence vanishing pairing; and $2\dim W=\dim Z$ because $Z$ is the direct sum of the pairs $V_i\oplus V_{4a-i}$ with $i<2a$ and $\dim V_i=\dim V_{4a-i}$ by [F4].

The off-middle blocks contribute nothing: by steps 2.2 and 2.3, $Z$ carries a nondegenerate symmetric form with the half-dimensional totally isotropic subspace $W$, so $\mathrm{sign}(Q_{M\times N}{\mathrm{\mid }}_Z)=0$ by [F5]. Moreover $Z\perp V_{2a}$ by step 1.2, and $V_{2a}$ is nondegenerate because $Q_{M\times N}$ is nondegenerate and $Z$ is nondegenerate with $H^{2k}(M\times N;\mathbb{R})=Z\oplus V_{2a}$; hence the inertia data of $Q_{M\times N}$ are the sums of those of $Q{\mathrm{\mid }}_Z$ and $Q{\mathrm{\mid }}_{V_{2a}}$.

Therefore $\sigma (M\times N)=\mathrm{sign}(Q{\mathrm{\mid }}_Z)+\mathrm{sign}(Q{\mathrm{\mid }}_{V_{2a}})=0+\mathrm{sign}(Q_M\otimes Q_N)=\sigma (M)\sigma (N)$, using step 3.1, step 2.1 and the multiplicativity of inertia under tensor products [F5]; the product smooth structure and orientation used are those of [F6] and the statement.