A particular complex of integral group ring is exact: proof of Jacobson

Let $$A$$ be a $$G$$-module.

Let $$C_n=\mathbb{Z}G\otimes \cdots \otimes \mathbb{Z}G$$ ($$n+1$$ copies). It is free $$\mathbb{Z}$$-module with basis $$g_0\otimes g_1\otimes \cdots \otimes g_n$$, $$g_i\in G$$.

Denote $$1\otimes g_1\otimes \cdots \otimes g_n\in C_n$$ by $$(g_1,\ldots,g_n)$$. These $$n$$-tuples are basis of free $$\mathbb{Z}G$$-module $$C_n$$.

Define $$\mathbb{Z}G$$-module homomorphisms $$d_n:C_n\rightarrow C_{n-1}$$ by $$d_n(g_1,\ldots,g_n)=g_1(g_2,\ldots,g_n)+\sum_{1}^{n-1}(-1)^i(g_1,\ldots, g_{i-1}, g_ig_{i+1},g_{i+2},\ldots, g_n)+(-1)^n(g_1,\ldots,g_{n-1}).$$ Let $$\varepsilon:\mathbb{Z}G(=C_0)\rightarrow \mathbb{Z}$$ be augmentation map.

Claim: The complex $$\mathbf{(C)}$$: $$\cdots \rightarrow C_2\xrightarrow{d_2}C_1\xrightarrow{d_1}C_0\xrightarrow{\varepsilon} \mathbb{Z}\rightarrow 0$$ is exact.

Proof in Jacobson's Algebra Vol.2 We define homomorphisms of abelian groups $$s_i:C_{i+1}\rightarrow C_i$$ such that $$\varepsilon s_{-1}=1_{\mathbb{Z}}; d_1s_0+s_{-1}\varepsilon=1_{C_0}; d_{n+1}s_n + s_{n-1}d_n=1_{C_n}.$$

Question: These conditions say that the maps $$s_i$$ define a homotopy between identity morphism $$(\mathbf{C})\rightarrow (\mathbf{C}$$) is homotopic to zero morphism from $$(\mathbf{C})\rightarrow (\mathbf{C}$$). But I didn't get how this implies the exactness of $$(\mathbf{C})$$? This could be trivial, but I was unable to see clearly the justification of Jacobson.

Suppose $$d_n(a)=0$$ ($$a$$ is a cycle). Then $$a=(d_{n+1}s_n+s_{n-1}d_n)(a)=(d_{n+1}s_n)(a)=d_{n+1}(b)$$ for $$b=s_n(a)$$. Thus $$a$$ is a boundary. All cycles are boundaries, so the chain complex is exact.