In "Differential forms in algebraic topology" by Bott & Tu, I am asked to prove the Kunneth Formula for singular cohomology by spectral sequence:

Exercise 15.12 (Kunneth Formula for Singular Cohomology). If $X$ is a space having a good cover, e.g., a triangularizable space, and $Y$ is any topological space, prove using the spectral sequence of the fiber bundle $\pi:X\times Y\rightarrow X$ that
\begin{eqnarray}
H^{n}(X\times Y)=\oplus\_{p+q=n}H^p(X,H^q(Y)).
\end{eqnarray}

The following is where I have arrived. Let us say that the coefficient ring is $\mathbb{Z}$. By the triviality of the bundle, I have reached that the second page is over a constant presheaf and the spectral sequence stops at the second page
\begin{eqnarray}
E^{p,q}\_\infty\cong E^{p,q}\_2=H^p(X,H^q(Y)),
\end{eqnarray}
with
\begin{eqnarray}
d\_2=d\_3=\cdots=0.
\end{eqnarray}
However, it only means that the spectral sequence converges to $H^\*(X\times Y)$, i.e.,
\begin{eqnarray}
H^n(X\times Y)&=&F\_0\supset F\_1\supset F\_2\supset\cdots,\\
F\_i/F\_{i+1}&\cong& H^i(X,H^{n-i}(Y)).
\end{eqnarray}
If the coefficient ring were a field, then the proof can be readily done. Since the coefficient ring now is $\mathbb{Z}$, there can be potentially an extension issue. I don't know how to deal with that. Or should I use $d\_2=d\_3=\cdots=0$ to prove that there is no such issues?