Assume ($M_1,\omega_1 $) and ($M_2,\omega_2)$ are symplectic manifolds. Consider $M_1\times M_2$. I want to show this is also symplectic manifold.
Let $p_1, p_2$ denote the projections. Then what I need is $p_1^*(\omega_1)^n \wedge p_2^*(\omega_2)^n$ is non-zero. I am stuck here. I guess $p_1^*(\omega_1 ^n)$ and $p_2^*(\omega_2 ^n)$ are both nonzero since $\omega_i ^n$ is nonzero by non degeneracy of $\omega_i$ and $p_i$ is submersion. But how do I know $p_1^*(\omega_1)^n \wedge p_2^*(\omega_2)^n$ is non-zero?
If $(M_1,\omega_1)$ and $(M_2,\omega_2)$ are symplectic manifolds then $$ \omega:=p_1^\ast\omega_1+p_2^\ast\omega_2 $$ is a symplectic form on $M_1\times M_2$. Then $\omega$ is closed since $d$ commutes with $p_1^\ast$ and $p_2^\ast$ and $\omega_1$ and $\omega_2$ are closed.
To see that $\omega$ is nondegenerate, let $x:=(x_1,x_2)\in M_1\times M_2$ and let $(u_1,u_2)\in T_x(M_1\times M_2)$ be a nonzero element. WLOG assume $u_1\neq 0$. Then there exists $v_1\in T_{x_1}M_1$ such that $\omega_1(u_1,v_1)\neq 0$. So $$ \omega((u_1,u_2),(v_1,0))=\omega_1(u_1,v_1)+\omega(u_2,0)=\omega_1(u_1,v_1)\neq 0 $$