This comes from Hatcher's Algebraic Topology book on page 139
He says that the map in the diagram
$$0 \to H_n(X^n) \stackrel{j_n}\to H_n(X^n, X^{n-1}) \stackrel{\partial_n}\to H_{n-1}(X^{n-1})\to?$$
The map $d_nd_{n+1}= j_{n+1}(\partial _nj_n)\partial_{n+1} = 0$ because of $\partial _nj_n = 0$ due to exactness. I don't know which exact sequence he is referring to in his diagram
Since $(X^n, X^{n-1})$ is a good pair, it induces the long exact sequence
$0 \to H_n(X^n) \to H_n(X^n, X^{n-1}) \to H_{n-1}(X^{n-1}) \to H_{n-1}(X^n) \to \dots$
Definition of exactness leads to $img j_n = \ker \partial_n$, so $\partial_n(img j_n) = \partial_n(\ker \partial_n) = 0 $