I'm trying to prove that if a map $f \colon X \to Y$ induces isomorphisms on singular homology with coefficients in $\mathbb{Z}/n\mathbb{Z}$, then the same is true for coefficients in $\mathbb{Z}/n^k\mathbb{Z}$ for all $k$.
Attempt: So let $\text{cone}(f)$ be the mapping cone of $f$, then $f$ is an $R$-homology isomorphism for a ring $R$ if and only if $\widetilde{H}_*(\text{cone}(f);R) = 0$.
The exact sequence $0 \to \mathbb{Z} \xrightarrow{\cdot n} \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z} \to 0$ of abelian groups induces (see Bockstein homomorphism) a long exact sequence
$$ \ldots \to H_n(\text{cone}(f);\mathbb{Z}) \to H_n(\text{cone}(f);\mathbb{Z}) \to H_n(\text{cone}(f);\mathbb{Z}/n\mathbb{Z}) \to H_{n-1}(\text{cone}(f);\mathbb{Z}) \to \ldots $$
Now the idea is to show that:
$\widetilde{H}_*(\text{cone}(f);\mathbb{Z}/n\mathbb{Z}) = 0 \iff \forall n : H_n(\text{cone}(f);\mathbb{Z}) \to H_n(\text{cone}(f);\mathbb{Z})$ above is iso.
If the map $\mathbb{Z} \xrightarrow{\cdot n} \mathbb{Z}$ induces isomorphisms in homology, then so does $\mathbb{Z} \xrightarrow{\cdot n^k} \mathbb{Z}$.
Need help here: Then the statement would pretty much follow. But my problem with 1. is that I don't see how it works in degree $n=0$, and for 2. I tried to use the naturally of the long exact sequence above combined with several maps between the sequences $$0 \to \mathbb{Z} \xrightarrow{\cdot n} \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z} \to 0$$ and $$0 \to \mathbb{Z} \xrightarrow{\cdot n^k} \mathbb{Z} \to \mathbb{Z}/n^k\mathbb{Z} \to 0$$ and I tried to use the five lemma to conclude 2., but it doesn't work out whatever maps I'm using.
Use induction, the 5-Lemma and the bockstein sequence coming from $0 \to \mathbb{Z}/n\mathbb{Z}\to \mathbb{Z}/n^k\mathbb{Z} \to \mathbb{Z}/n^{k-1}\mathbb{Z} \to 0$.