Homology Problem I'm having trouble with the following two part problem (for self studying).

Consider the $n$-simplex $\Delta^n$ together with all its faces as a
$\Delta$-complex; it has a simplex for every non-empty subset of
$\{1,...,n\}$. Show that $H_{0}^{\Delta} (\Delta^n) = \mathbb{Z}$ and
$H_{i}^{\Delta}(\Delta^n)=0$ for all $i>0$.

I'm thinking of considering maps $f_{i}:\Delta_{i}(\Delta^n)\rightarrow \Delta_{i+1}(\Delta^n)$ that send the simplex $[v_0,...,v_i]\rightarrow [0,v_0,...,v_i]$ if $v_0 \neq 0$ and $0$ otherwise.

The part two of this question is that if $n>1$, consider the
$\Delta$-complex $\partial \Delta^n$ which has as simplices all faces
of $\Delta^n$ except $\Delta^n$ itself, so it has a simplex for every
non-empty subset of $\{0,1,...,n\}$. Show that
$H_{0}^{\Delta}(\partial \Delta^n) \cong \mathbb{Z}$ and
$H_{n-1}^{\Delta}(\partial \Delta^n) \cong \mathbb{Z}$ and all other
simplicial homology groups vanish.

I'm pretty sure that part two uses part one since it's almost the same problem except we're cutting out $\Delta^n$. I'm dead stuck. How do I make the proof with the necessary steps? Thanks in advance.
 A: Goal: Let's find the homology groups associated to the simplicial complexes:
$$\Delta^n\text{ }\text{ }\text{ and }\text{ }\text{ }\partial\Delta^n.$$

Read these for additional background:

Simplex | Simplicial Complex | Simplicial Homology

$\bullet$ $\underline{\text{Def:}}$ In $\mathbb{R}^{n+1}$, take a set of Linearly-Independent vectors,
$$S:= \big\{v_1,...,v_n\big\},$$
then tack on an origin point, $v_0\in \mathbb{R}^{n+1}$, that is mutually L.I. with this set, $S = S\cup \{v_0\}$. We may write an expression for a $\color{red}{\text{simplex based at }v_0}$ as:
$$\color{red}{\sigma} := Span\big\{S\big\}\bigg|_{\big\{v\text{ }\text{ }\big|\text{ }\sum\limits_{i=0}^{n+1}(v)^i = 1\big\}}$$


Next,
$\bullet$ $\underline{\text{Def:}}$ A $\color{red}{\text{simplicial complex}}$ is a set of simplices:
$$\color{red}{K} := \bigg\{\sigma\text{ }|\text{ }\sigma \text{ is a simplex}\bigg\}$$
that is closed under restriction to sub-simplices (generated by subsets of vertices):
$$\forall \sigma \in K, \forall\tau\leq \sigma:\text{ }\tau\in K$$
$\underline{\text{Note:}}$ We may now define the complexes:
$$\color{blue}{\Delta^n} = \langle \sigma \rangle_{\leq}\text{ }\text{ }\text{ and }\text{ }\text{ }\color{blue}{\partial \Delta^n} = \Delta^n - \sigma,$$
where $\sigma = (v_0,...,v_n)$ also indicates orientation with parenthesis.

$\bullet$ $\underline{\text{Def:}}$ The $\color{red}{\text{set of }m\text{-chains in K}}$ is given by the (finite) formal linear combinations of (oriented) $m$-simplices:
$$\color{red}{C_m} := \bigg\{\sum\limits_{i\in I\\ |I|<\infty}\lambda_i\sigma_i\text{ }\bigg|\text{ }\sigma_i\in K\text{ and }\lambda_i\in\mathbb{Z}\bigg\}$$
$\bullet$ $\underline{\text{Def:}}$ The $\color{red}{m^{th}\text{-boundary operator}}$ is then given by:
$$\color{red}{\partial_m} :C_m \to C_{m-1}$$
$$\bigg(\sum\limits_{i\in I}\lambda_i\sigma_i\bigg)\mapsto \bigg(\sum\limits_{i\in I}\lambda_i\partial_m(\sigma_i)\bigg)$$
where atomically on oriented $m$-simplices we have:
$$\partial_m (\sigma) = \partial_m (v_0,...,v_m) := \sum\limits_{j=0}^m(-1)^j*(v_0,...,\hat{v}_j,...,v_m).$$
The negative is contrived for orientation purposes and hat is removal as usual.

$\bullet$ $\underline{\text{Def:}}$ Now, with the chain complex, $\partial^2 = 0$ (glossing over alot), we have $\color{red}{\text{homology groups}}$ as quotients of subgroups in the nodes. Specific to $\Delta^n$, we have:

$$\color{red}{H_m} := ker(\partial_{m})/Im(\partial_{m+1}) $$
$$ = \bigg\{\bigg\{\sum\limits_{i_1\in I_1}\lambda_{i_1}\sigma_{i_1}+\sum\limits_{i_2\in I_2}\lambda_{i_2}\partial_{m+1}(\sigma_{i_2})\text{}\bigg|\text{ where }\partial_{m}(\sigma_{i_1})=0,\text{ }\lambda_{i_a}\in\mathbb{Z},\text{ and }\sigma_{i_a}\in K\bigg\}_{\text{fixed first term}}\bigg\}$$
#setOfCosets
From here the result should be easier to get to.
There are no $n+1$ or $-1$ symplices in our $K$, hence the initial and terminal nodes are empty, together with trivial inclusion map, $\partial_{n+1}$, and the zero map, $\partial_0$.


$\underline{\text{Calculating }H_0(\Delta^n):}$
$$H_0 = ker(\partial_0)/Im(\partial_1).$$
But $ker(\partial_0) = C_0$. And with images of $\partial_1$ being of the atomic form:
$$\partial_1((v^0_j,v^1_j)) = v^1_j-v^0_j$$
we get:
$$H_0 = \bigg\{\bigg\{\sum\limits_{i_1\in I_1}\lambda_{i_1} (v_{i_1})+\sum\limits_{i_2\in I_2}\lambda_{i_2}(v^1_{i_2}-v^0_{i_2})\text{}\bigg|\text{ }\lambda_{i_a}\in\mathbb{Z}\text{ and }v^k_{i_a}\in \sigma\bigg\}_{\text{fixed first term}}\bigg\}$$
Note we may reduce this to the set of equivalence classes of the form:
$$[\lambda (v_0)]\text{ for }\lambda\in \mathbb{Z}$$
Hence $\color{blue}{H_0 \cong \mathbb{Z}}$.

Moving towards the others, $H_1(\Delta^n) = ker(\partial_1)/Im(\partial_2)$...

