Explicit calculation of simplicial homology Is it possible to calculate simplicial homology of $n$-dimensional simplex just by definition, without using homotopy invariance of homology(or it's equality to singular or cellular ones)?
I've done it for $n\leq4$ by explicit calculation of $\partial$, but this method doesn't seems to be easy generalizable.
UPD: I'm considering strucuture of simplicial complex on simplex there simplices of complex are all faces of simplex. I'm using this definition of simplicial homology.
 A: HINT:
Yes you can. You won't mention the homotopy invariance, but  you will use a contraction.
Say you have a contractible space $X$, like your simplex. It comes with a map
$$\phi\colon X\times [0,1]\to X$$
$\phi_{|X\times \{1\}} = \text{Id}_X$ and $\phi_{|X\times \{0\}} \equiv x_0$
( $X$ contracts to the point  $x_0$ ). 
Using this we'll construct a map from singular $n$-simplexes on $X$ to singular $n+1$-simplexes on $X$ as follows:
Let $f\colon \Delta_n \to X$ a singular simplex. This gets us a map 
$$f \times \text{Id}_I \colon \Delta_n \times I \to X\times I $$
Compose it with $\phi$ to get a map from $\Delta_n \times I$ to $X$ that takes the value $x_0$ on $\Delta_n \times \{0\}$, so it factors through $\Delta_n \times I/\Delta_n \times \{0\}$. But $\Delta_n \times I/\Delta_n \times \{0\}$ homeomorphic to $\Delta_{n+1}$ in a natural way. We got ourselves a singular $n+1$-simplex.
In this way we get a map of complexes $\tau_n\colon C_{n}(X) \to C_{n+1}(X)$ for all $n\ge 0$. 
This allows us to show that the (reduced) homology is zero. Hint: calculate
$$d \tau + \tau d$$ in each dimension. 
