Trying to prove $\partial^2=0$ on $k$-cells Consider a $k$-cell $ c : [0,1]^k \to U \subset \mathbb{R}^n , (t_1,...,t_k) \mapsto c (t_1,...,t_k) $. Then the boundary of $c$, $\partial c$ is defined as
$$ \partial c := \sum_{i=1}^k (-1)^{i-1} \left( c_i^1 - c_i^0 \right) $$
where 
$$ c_i^1 := c(t_1,...,t_{i-1},1,t_{i+1},...,t_k)  \\ c_i^0 := c(t_1,...,t_{i-1},0,t_{i+1},...,t_k) $$ 
Then supposing $c = \sum_m a^m c_m$ is a $k$-chain ($a^m \in \mathbb{R}$, $c_m$ $k$-cells), it’s boundary is defined by
$$ \partial c = \partial \left( \sum_m a^m c_m \right) := \sum_m a^m \partial c_m $$ 
I’m struggling to prove the special property of $\partial$ that $\partial^2 = 0$. Here’s what I have so far... 
Since $\partial$ is linear, it is sufficient to prove $\partial^2 c = 0$ for any $k$-cell $c$. 
Introducing another sigma sign, the definition of $\partial c$ can be made to look nicer,  
$$ 
\partial c := \sum_{i=1}^k (-1)^{i-1} \left( c_i^1 - c_i^0 \right) 
\\ 
=  \sum_{i=1}^k (-1)^{i-1} \sum_{\rho=0}^1 (-1)^{\rho+1} c_i^\rho   \\
= \sum_{i=1}^k \sum_{\rho=0}^1 (-1)^{i+\rho} c_i^\rho
$$
Then, I tried brute forcing some algebra and definitions... 
$$
\partial^2 c = \partial \sum_{i=1}^k \sum_{\rho=0}^1 (-1)^{i+\rho} c_i^\rho 
= \sum_{i=1}^k \sum_{\rho=0}^1 (-1)^{i+\rho} \partial c_i^\rho
\\ 
= \sum_{i=1}^k \sum_{\rho=0}^1 (-1)^{i+\rho} \sum_{j=1 | j\neq i}^k \sum_{\sigma=0}^1 (-1)^{j+\sigma} c_{ij}^{\rho \sigma} 
\\ 
= \sum_{i,j=1 | i\neq j }^{k} \sum_{\rho,\sigma = 0}^{1} (-1)^{i+j+\rho+\sigma} c_{ij}^{\rho\sigma} 
$$
where 
$$
c_{ij}^{\rho\sigma} = c(t_1, ... , t_{i-1} , \rho , t_{i+1} , ... , t_{j-1} , \sigma, t_{j+1}, ... , t_k) 
$$
This is where I’m stuck. 
 A: Note that if $c$ is a $k$-cell, i.e. $c : [0,1]^k \to \mathbb R^n$ then $c_i^\rho$ is a $(k-1)$-cell, and — here is the crux — the first $i-1$ argument positions coincide with those of $c$, but the ones after the $i$th position don't; they are shifted by one step. Therefore,
$$\partial c_i^\rho = \sum_{\underset{i\neq j}{j=1,\ldots,k}} (-1)^{j+[j>i]+1} (c_{ij}^{\rho 1} - c_{ij}^{\rho 0}),$$
where $[j>i] = 1$ when $j>i$ and $= 0$ otherwise.
Your big sum then is symmetric in $i$ and $j$ except for the factor $(-1)^{[j>i]}$ which makes all terms cancel:
$$
\partial^2 c 
= \sum_{\underset{i \neq j}{i,j=1,\ldots,k}} (-1)^{i+j+[j>i]+2} (c_{ij}^{11} - c_{ij}^{10} - c_{ij}^{01} + c_{ij}^{00}) \\
= \underbrace{\sum_{\underset{i \neq j}{i,j=1,\ldots,k}}}_{\text{symmetric in $i,j$}} \underbrace{(-1)^{[j>i]}}_{\text{antisymmetric in $i,j$}} \underbrace{ (-1)^{i+j+2} (c_{ij}^{11} - c_{ij}^{10} - c_{ij}^{01} + c_{ij}^{00})}_{\text{symmetric in $i,j$}}
= 0$$
A: You approach is absolutely correct. Let us do it a little bit more precise.
For $\rho = 0,1$ and $i = 1,...,k$ let us define maps
$$s_{i,k}^\rho: I^{k-1} \to I^k, s_{i,k}^\rho(x_1,..., x_{k-1}) = (x_1,...,x_{i-1},\rho,x_i,...,x_{k-1}) .$$
Then for $i \le j$
$$s_{i,k}^\rho \circ s_{j,k-1}^\sigma = s_{j+1,k}^\sigma \circ s_{i,k-1}^\rho  .$$
We have $c_i^\rho = c \circ s_{i,k}^\rho$. Define $P = \{1,..., k \} \times \{1,...,k-1 \}$, $A = \{ (i,j) \in P \mid i > j \}$ and $B = \{ (i,j) \in P \mid i \le j \}$. We have $(i,j) \in B$ if and only if $(j+1,i) \in A$.
As you computed
$$
\partial^2 c = \sum_{i=1}^k \sum_{\rho=0}^1 (-1)^{i+\rho} \partial c_i^\rho
\\ 
= \sum_{i=1}^k \sum_{\rho=0}^1 (-1)^{i+\rho} \sum_{j=1 }^{k-1} \sum_{\sigma=0}^1 (-1)^{j+\sigma} (c_i^\rho)_j^\sigma 
\\ 
= \sum_{(i,j) \in P} \sum_{\rho,\sigma = 0}^{1} (-1)^{i+j+\rho+\sigma} c \circ s_{i,k}^\rho \circ s_{j,k-1}^\sigma
\\
= \sum_{(i,j) \in A} \sum_{\rho,\sigma = 0}^{1} (-1)^{i+j+\rho+\sigma} c \circ s_{i,k}^\rho \circ s_{j,k-1}^\sigma + \sum_{(i,j) \in B} \sum_{\rho,\sigma = 0}^{1} (-1)^{i+j+\rho+\sigma} c \circ s_{i,k}^\rho \circ s_{j,k-1}^\sigma
\\
= \sum_{(i,j) \in A} \sum_{\rho,\sigma = 0}^{1} (-1)^{i+j+\rho+\sigma} c \circ s_{i,k}^\rho \circ s_{j,k-1}^\sigma + \sum_{(i,j) \in B} \sum_{\rho,\sigma = 0}^{1} (-1)^{i+j+\rho+\sigma} c \circ s_{j+1,k}^\sigma \circ s_{i,k-1}^\rho
\\
= \sum_{(i,j) \in A} \sum_{\rho,\sigma = 0}^{1} (-1)^{i+j+\rho+\sigma} c \circ s_{i,k}^\rho \circ s_{j,k-1}^\sigma + \sum_{(m,n) \in A} \sum_{\alpha,\beta = 0}^{1} (-1)^{m+n-1+\alpha+\beta} c \circ s_{m,k}^\alpha \circ s_{n,k-1}^\beta
\\
= 0
$$
where we used $m = j+1, n = i$.
