= Solution
Write $\rho:H^*(X;\mathbb Z)\to H^*(X;\mathbb Z/n)$ for coefficient reduction. Compare the two coefficient sequences in part (a): the maps from the integral sequence to the finite sequence are reduction modulo $n$ on the left, reduction modulo $n^2$ in the middle, and the identity on the right. The square involving the injections commutes because $[na]_{n^2}=j([a]_n)$.
Naturality of the <connecting homomorphism> gives the <Bockstein factorization through integral cohomology>
$$
\boxed{\beta=\rho\widehat\beta.}
$$
One can see this directly without a diagram: lift a modulo-$n$ cocycle to an integral cochain $a$. Its coboundary has the form $\delta a=nb$. Then $\widehat\beta[a]=[b]$, whereas $\beta[a]=[b\bmod n]$.
Exactness of the integral coefficient sequence gives $\widehat\beta\rho=0$: a class obtained by reducing an integral cocycle has zero integral connecting class. Consequently
$$
\boxed{\beta^2=\rho\widehat\beta\rho\widehat\beta=0.}
$$
This <Bockstein square-zero identity> holds without requiring $n$ to be prime. At the cochain level, $\delta^2a=0$ and the torsion-free integral cochain groups imply $\delta b=0$, which also makes the second connecting class vanish.
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