face map $d_i^n:\Delta^{n-1}\to\Delta^n$ I have the following formulas about face maps, which I try to proof.

1) $d_j^{n+1}d_i^n=[e_0,\dotso, \hat{e_i},\dotso, \hat{e_j},\dotso, e_{n+1}]:\Delta^{n-1}\to\Delta^{n+1}$ for $j>i$.
2) $d_j^{n+1}d_i^n=[e_0,\dotso,\hat{e_j},\dotso, \hat{e_{i+1}},\dotso, e_{n+1}]:\Delta^{n-1}\to\Delta^{n+1}$ for $j\leq i$
3) $d_j^{n+1}d_i^n=d_{i+1}^{n+1}d_j^n$ for $j\leq i$

Proof:
Let $(x_0,\dotso, x_{n-1})\in\Delta^{n-1}$. Then is
$d_j^{n+1}(d_i^n(x_0,\dotso, x_{n-1}))=d_j^{n+1}(x_0,\dotso, x_{i-1}, 0,\color{red}{x_i}, x_{i+1},\dotso, x_{n})=(x_0,\dotso, x_{i-1}, 0, \color{red}{x_i},x_{i+1},\dotso, x_{j-1}, 0,\color{red}{x_j}, x_{j+1},\dotso, x_{n+1})$
The second formula can be proven similary, where in the case of $i=j$ the final result is a vector which looks like this:
$(x_0, \dotso, x_{i-1}, 0, 0, x_{i+2},\dotso, x_{n+1})$
Am I right?
The 3rd formula then just follows from the first two.
Thanks in advance.

Definition: For $0\leq i\leq n-1$ is the $i-th$ face map defined as $d_i^n:\Delta^{n-1}\to\Delta^n$ induced by $[e_0,\dotso, \hat{e_i},\dotso, e_n]:\Delta^{n-1}\to\mathbb{R}^{n+1}$

 A: I know now what is happening, and that is why one should declare what notations mean to avoid confusion.  In this answer, I attempt to fix the errors and undefined terms in your question.  As usual, the $k$-dimensional simplex is the set $\Delta^k\subseteq\mathbb{R}^{n+1}$ defined by
$$\Delta^k:=\Big\{\left(x_0,x_1,x_2,\ldots,x_k\right)\in\mathbb{R}^{k+1}\,\Big|\,x_0,x_1,x_2,\ldots,x_k\geq 0\text{ and }x_0+x_1+x_2+\ldots+x_k=1\Big\}\,.$$
First of all, $e_0,e_1,e_2,\ldots,e_n$ are standard basis vectors of $\mathbb{R}^{n+1}$ (and by abuse of notation, $e_0,e_1,e_2,\ldots,e_{n+1}$ are also standard basis vectors of $\mathbb{R}^{n+2}$).  Then, you try to identify $\Delta^{n-1}$ with the convex combination $$\left[e_0,e_1,e_2,\ldots,e_{i-1},e_{i+1},e_{i+2},\ldots,e_n\right]=\left[e_0,e_1,\ldots,\hat{e}_i,\ldots,e_n\right]\,.$$
Then, the map $d_i^n:\Delta^{n-1}\to \Delta^n$ is defined to be
$$d_i^n\left(x_0,x_1,x_2,\ldots,x_{n-1}\right):=\left(x_0,x_1,\ldots,x_{i-1},0,x_i,x_{i+1},\ldots,x_{n-1}\right)\in \Delta^n$$ for all $\left(x_0,x_1,x_2,\ldots,x_{n-1}\right)\in\Delta^{n-1}$.  This means the image of $d_i^n$ is precisely $\left[e_0,e_1,\ldots,\hat{e}_i,\ldots,e_n\right]$. That is, $d_i^n$ is the embedding of $\Delta^{n-1}$ into one $(n-1)$-dimensional face of $\Delta^n$.   (I don't know why you wrote something like $d_j^{n+1}\circ d_i^{n}=\left[e_0,e_1,\ldots,\hat{e}_i,\ldots,\hat{e}_j,\ldots,e_{n+1}\right]$.  This makes no sense.  A function does not equal a simplex.)  
Hence, it is easy to check all the statements.  Let $\left(x_0,x_1,x_2,\ldots,x_{n-1}\right)$ be an arbitrary point of $\Delta^{n-1}$.  
Now, we shall deal with Part (1).  For $j>i$, 
$$\begin{align}\left(d_{j}^{n+1}\circ d_i^{n}\right)\left(x_0,x_1,x_2,\ldots,x_{n-1}\right)&=d_j^{n+1}\left(x_0,x_1,\ldots,x_{i-1},0,x_i,x_{i+1},\ldots,x_n\right)\\&=\left(x_0,x_1,\ldots,x_{i-1},0,x_i,\ldots,x_{j-2},0,x_{j-1},x_j,\ldots,x_{n-1}\right)\end{align}$$
has zeros in the $i$-th and $j$-th coordinates.  That is, $d_j^{n+1}\circ d_{i}^{n}:\Delta^{n-1}\to \Delta^{n+1}$ satisfies $$\text{im}\left(d_j^{n+1}\circ d_i^{n}\right)=\left[e_0,e_1,\ldots,\hat{e}_i,\ldots,\hat{e}_j,\ldots,e_{n+1}\right]\,,$$
as required.
Next, we tackle Part (2).  For $j\leq i$,
$$\begin{align}\left(d_{j}^{n+1}\circ d_i^{n}\right)\left(x_0,x_1,x_2,\ldots,x_{n-1}\right)&=d_j^{n+1}\left(x_0,x_1,\ldots,x_{i-1},0,x_i,x_{i+1},\ldots,x_n\right)\\&=\left(x_0,x_1,\ldots,x_{j-1},0,x_j,\ldots,x_{i-1},0,x_{i},x_{i+1},\ldots,x_{n-1}\right)\end{align}$$
has zeros in the $j$-th and $(i+1)$-st coordinates.  That is, $d_j^{n+1}\circ d_{i}^{n}:\Delta^{n-1}\to \Delta^{n+1}$ satisfies $$\text{im}\left(d_j^{n+1}\circ d_i^{n}\right)=[e_0,e_1,\ldots,\hat{e}_j,\ldots,\hat{e}_{i+1},\ldots,e_{n+1}]\,.$$
Finally, we justify the equality in Part (3).  For $j\leq i$, we see that, $j<i+1$.  Therefore, from Part (1),
$d_{i+1}^{n+1}\circ d_j^{n}:\Delta^{n-1}\to\Delta^{n+1}$ satisfies $$\text{im}\left(d_{i+1}^{n+1}\circ d_j^{n}\right)=[e_0,e_1,\ldots,\hat{e}_j,\ldots,\hat{e}_{i+1},\ldots,e_{n+1}]=\text{im}\left(d_j^{n+1}\circ d_i^{n}\right)\,.$$  Additionally,
$$\begin{align}\left(d_{i+1}^{n+1}\circ d_j^{n}\right)\left(x_0,x_1,x_2,\ldots,x_{n-1}\right)&=\left(x_0,x_1,\ldots,x_{j-1},0,x_j,\ldots,x_{i-1},0,x_{i},x_{i+1},\ldots,x_{n-1}\right)\\&=\left(d_{j}^{n+1}\circ d_i^{n})(x_0,x_1,x_2,\ldots,x_{n-1}\right)\end{align}$$
also by Part (1).  This proves that $d_{i+1}^{n+1}\circ d_j^{n}=d_{j}^{n+1}\circ d_i^{n}$.
