[proofplan] The proof is a direct unpacking of the definition of homology. First we use the chain complex identity $d_n \circ d_{n+1} = 0$ to show that every boundary in degree $n$ is a cycle, so the quotient $\ker(d_n)/\operatorname{im}(d_{n+1})$ is defined. Then we prove that this quotient module is the zero module exactly when its denominator equals its numerator. [/proofplan] [step:Identify cycles and boundaries in degree $n$] Define the $R$-submodule of $n$-cycles by \begin{align*} Z_n(C_\bullet) := \ker(d_n) \subset C_n, \end{align*} and define the $R$-submodule of $n$-boundaries by \begin{align*} B_n(C_\bullet) := \operatorname{im}(d_{n+1}) \subset C_n. \end{align*} Since $C_\bullet$ is a chain complex, $d_n \circ d_{n+1} = 0$. Therefore, for every $x \in C_{n+1}$, \begin{align*} d_n(d_{n+1}(x)) = 0, \end{align*} so $d_{n+1}(x) \in \ker(d_n)$. Hence \begin{align*} B_n(C_\bullet) \subset Z_n(C_\bullet). \end{align*} Thus the quotient module \begin{align*} H_n(C_\bullet) := Z_n(C_\bullet)/B_n(C_\bullet) \end{align*} is well-defined. [/step] [step:Show that vanishing homology forces exactness at $C_n$] Assume \begin{align*} H_n(C_\bullet) = 0. \end{align*} By the definition of homology, this means \begin{align*} Z_n(C_\bullet)/B_n(C_\bullet) = 0. \end{align*} Let $z \in Z_n(C_\bullet)$. The coset $z + B_n(C_\bullet)$ is an element of the zero quotient module, so it equals the zero coset: \begin{align*} z + B_n(C_\bullet) = B_n(C_\bullet). \end{align*} Therefore $z \in B_n(C_\bullet)$. Since $z \in Z_n(C_\bullet)$ was arbitrary, we have \begin{align*} Z_n(C_\bullet) \subset B_n(C_\bullet). \end{align*} Together with $B_n(C_\bullet) \subset Z_n(C_\bullet)$ from the previous step, this gives \begin{align*} Z_n(C_\bullet) = B_n(C_\bullet). \end{align*} Substituting the definitions of $Z_n(C_\bullet)$ and $B_n(C_\bullet)$, we obtain \begin{align*} \ker(d_n) = \operatorname{im}(d_{n+1}), \end{align*} so the complex is exact at $C_n$. [/step] [step:Show that exactness at $C_n$ forces vanishing homology] Assume that $C_\bullet$ is exact at $C_n$, so \begin{align*} \operatorname{im}(d_{n+1}) = \ker(d_n). \end{align*} Using the definitions of $Z_n(C_\bullet)$ and $B_n(C_\bullet)$, this says \begin{align*} B_n(C_\bullet) = Z_n(C_\bullet). \end{align*} Hence \begin{align*} H_n(C_\bullet) = Z_n(C_\bullet)/B_n(C_\bullet) = Z_n(C_\bullet)/Z_n(C_\bullet) = 0. \end{align*} Thus $H_n(C_\bullet)=0$. [/step] [step:Conclude the equivalence] We have shown that $H_n(C_\bullet)=0$ implies \begin{align*} \operatorname{im}(d_{n+1}) = \ker(d_n), \end{align*} and that this equality implies $H_n(C_\bullet)=0$. Therefore $H_n(C_\bullet)=0$ if and only if $C_\bullet$ is exact at $C_n$. [/step]