= Solution
For <chain maps> $f,g:C_*\to D_*$, a <chain homotopy> from $g$ to $f$ is a family of <group homomorphisms> $s_q:C_q\to D_{q+1}$ satisfying
$$
\boxed{f_q-g_q=d^D_{q+1}s_q+s_{q-1}d^C_q\quad\text{for every }q.}
$$
If $z$ is a <chain cycle>, then $(f-g)(z)=d^Ds(z)$ is a <chain boundary>. Consequently $[f(z)]=[g(z)]$, so $\boxed{f_*=g_*}$ on every <homology group>.
Order the vertices of the <simplex> as $v_0,\ldots,v_n$. Its <simplicial chain complex> has
$$
S_q(\Delta^n)=\bigoplus_{0\leq i_0<\cdots<i_q\leq n}\mathbb Z[v_{i_0},\ldots,v_{i_q}]
\quad(0\leq q\leq n),
$$
and zero groups otherwise. An arbitrary ordering represents the sign of the <permutation> times the increasing ordering. Its <boundary operator> is
$$
d_q[v_{i_0},\ldots,v_{i_q}]
=\sum_{r=0}^q(-1)^r[v_{i_0},\ldots,\widehat v_{i_r},\ldots,v_{i_q}]
\quad(q\geq1),\qquad d_0=0.
$$
In $d^2$ each face obtained by deleting the vertices in positions $r<t$ occurs twice. Deleting $r$ first gives sign $(-1)^r(-1)^{t-1}$, whereas deleting $t$ first gives $(-1)^t(-1)^r$. These signs are opposite, so all terms cancel. The case $d_0d_1=0$ follows directly from $d_0=0$. Thus $\boxed{d^2=0}$.
For the <homology> calculation, add the augmentation $\varepsilon:S_0\to\mathbb Z$ sending every vertex to $1$. Regard $\mathbb Z$ as the group in degree $-1$ of an <augmented chain complex>. Define the <simplicial cone chain contraction> by
$$
s_{-1}(1)=[v_0],\qquad
s_q[v_{i_0},\ldots,v_{i_q}]=
\begin{cases}
[v_0,v_{i_0},\ldots,v_{i_q}],&i_0>0,\\
0,&i_0=0.
\end{cases}
$$
If the simplex does not contain $v_0$, expansion of its coned boundary gives $ds(\sigma)=\sigma-sd(\sigma)$. If it contains $v_0$, the only face on which $s$ is nonzero is the face omitting $v_0$, and $sd(\sigma)=\sigma$. In degree zero this uses the augmentation, and in degree $-1$ it says $\varepsilon s_{-1}=1$. Hence
$$
\boxed{ds+sd=1\quad\text{on the augmented complex}.}
$$
Every positive-degree <chain cycle> is therefore a <chain boundary>. In degree zero, the same identity says $z-\varepsilon(z)[v_0]=ds_0(z)$, so the augmentation induces an isomorphism $H_0\to\mathbb Z$, with inverse $1\mapsto[v_0]$. Thus, including $n=0$,
$$
\boxed{H_q(S_*(\Delta^n))\cong
\begin{cases}\mathbb Z,&q=0,\\0,&q>0.\end{cases}}
$$
This is a direct <chain homotopy> calculation and uses no result about <cellular homology>.
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