[proofplan] The chain complex condition first ensures that every boundary is a cycle, so the quotient $H_n(C_\bullet) = Z_n(C_\bullet)/B_n(C_\bullet)$ is well-defined in each degree. The quotient in degree $n$ vanishes exactly when its numerator and denominator agree, namely when $Z_n(C_\bullet) = B_n(C_\bullet)$. Replacing cycles and boundaries by their definitions gives $\ker d_n = \operatorname{im} d_{n+1}$, which is precisely exactness at $C_n$. [/proofplan] [step:Use the chain complex condition to place boundaries inside cycles] Fix $n \in \mathbb{Z}$. Since $d_n \circ d_{n+1} = 0$, every element of $\operatorname{im} d_{n+1}$ is sent to $0$ by $d_n$. Therefore \begin{align*} B_n(C_\bullet) = \operatorname{im} d_{n+1} \subset \ker d_n = Z_n(C_\bullet). \end{align*} Thus $H_n(C_\bullet) = Z_n(C_\bullet)/B_n(C_\bullet)$ is a well-defined quotient $R$-module. [/step] [step:Translate vanishing homology in one degree into equality of cycles and boundaries] For the fixed degree $n$, the quotient $R$-module $H_n(C_\bullet)$ is zero if and only if every coset in $Z_n(C_\bullet)/B_n(C_\bullet)$ is the zero coset. This holds if and only if every element of $Z_n(C_\bullet)$ belongs to $B_n(C_\bullet)$, that is, \begin{align*} H_n(C_\bullet) = 0 \quad \iff \quad Z_n(C_\bullet) = B_n(C_\bullet). \end{align*} Using the definitions of $Z_n(C_\bullet)$ and $B_n(C_\bullet)$, this is equivalent to \begin{align*} H_n(C_\bullet) = 0 \quad \iff \quad \ker d_n = \operatorname{im} d_{n+1}. \end{align*} [/step] [step:Apply the degreewise equivalence in every integer degree] The chain complex $C_\bullet$ is exact precisely when \begin{align*} \operatorname{im} d_{n+1} = \ker d_n \end{align*} for every $n \in \mathbb{Z}$. By the degreewise equivalence proved above, this condition holds for every $n \in \mathbb{Z}$ if and only if \begin{align*} H_n(C_\bullet) = 0 \end{align*} for every $n \in \mathbb{Z}$. Hence $C_\bullet$ is exact if and only if all of its homology modules vanish. [/step]