n-dimensional simplex $\Delta^n$: every cycle without boundary is a boundary This question is concerned with the evaluation of the homology groups $H_k(\Delta^n)=H_k(\Delta^n,\mathbb Z)$, where $\Delta ^n$ is a $n$-dimensional simplex. Since $\Delta^n$ is retractable, $H_k(\Delta^n)=\{0\}$ for all $k$.
Now, let's try to evaluate $H_k(\Delta^n)$ directly by simplicial structure on $\Delta^n$. Indeed, all subsimplexes of $\Delta^n$ forms a simplicial structure of $\Delta^n$. Let $C_n$ denote the free abelian group with generators being the simplexes in this simplicial structure. We then have the boundary operators $\partial_n$ defined in the usual way. We expect the homologies $H_k(\Delta^n)=\ker \partial_k/\text{Im}\partial _{k+1}$ to be trivial, so we want to show that all $k$-dimensional subcomplexes without boundary are the boundary of some $(k+1)$-dimensional subcomplex.
The fact written in bold doesn't seem to be very obvious to prove geometrically and directly from the property of simplexes. Is there any way to show that $\ker \partial_k=\text{Im}\partial _{k+1}$ by considering the simplicial structure? Can one compute $H_k(\Delta^n)$ directly by simplicial structure on $\Delta^n$?
In other words, the result written in bold sounds pretty non-trivial for me and it is not obvious to prove without referring to the homotopy invariance of $H_n$ and a few other theorems. I want to know if it has an easy and elementary proof.
 A: One may calculate this directly. 
The k-simplices of $\Delta^n$ are given by $(k+1)$-element subsets of $\{0, \cdots, n\}$. The boundary operator is given by $$\partial \{i_0, \cdots, i_k\} = \sum_{j=0}^k (-1)^j \{i_0, \cdots, \hat i_j, \cdots, i_k\},$$ where the hat denotes "drop this variable". 
Consider the operation $f(S) = \{0\} \cup S$ when $0 \not \in S$ and $f(S) = 0$ when $0 \in S$. Geometrically, we are taking the "cone towards 0"; every $k$-simplex that already contains 0 is "coned again" and doesn't contribute any $(k+1)$-simplices (because the cone is contains in the k-simplex itself); that's why we set $f(S) = 0$ in those cases.
I claim that $fd + df = 1$. In particular, if $c$ is a cycle, this says that $c = f(d(c)) + d(f(c)) = d(f(c))$, so that $c$ is a boundary. 
Now we simply need to check this term-by-term. 
First, suppose $0 \in S$. Then we have $f(S) = 0$, and hence $df(S) = 0$. At the same time, $$d(\{0, i_1, \cdots, i_k\}) = \{i_1, \cdots, i_k\} + \sum_{j=1}^k (-1)^j \{0,i_1, \cdots, \hat i_j, \cdots, i_k\},$$ so that $$f(d(\{0, i_1, \cdots, i_k\})) = \{0, i_1, \cdots, i_k\}.$$ Thus for basis elements $S$ with $0 \in S$ we see that $fd + df = 1$. 
Now suppose $0 \not \in S$; that is, $S = \{i_0, \cdots, i_k\}$ where $i_0 > 0$. Then
$$f(d(S)) = \sum_{j =0}^k (-1)^j \{0, i_0, \cdots, \hat i_j, \cdots, i_k\},$$ while $$d(f(S)) = S + \sum_{j=0}^k (-1)^{j+1} \{0, i_0, \cdots, \hat i_j, \cdots, i_k\}.$$ 
In particular we have that almost all terms of $(df+fd)(S)$ cancel out; we are left with $(df+fd)(S) = S$. Thus we have checked that $df + fd = 1$ on basis elements, so it is true for every chain, and as above we have this constructed a chain $f(c)$ so that for every cycle $c$, we have $d(f(c)) = c$.
