Let $c\_i$ be the number of complete subgraphs of size $i$ in a graph $G$. I learned the following result:


If $G$ has $n$ vertices and is connected and chordal, that is, $G$ does not have induced cycles of length at least 4, then 
$$
\sum\_{i=1}^n (-1)^{i+1} c\_i = 1
$$


See for example Thm 4.1 (together with the fact that $\lambda=1$ is a root after Theorem 3.1) in this [paper](http://www.sciencedirect.com/science/article/pii/S0012365X02005009), or Prop 2.3 of [this paper](https://arxiv.org/pdf/1004.3416.pdf).


My question is the following:


1. Is there an elementary proof of this fact without resorting to topological arguments?
2. Is there a generalization of this to k-chordal graphs?



Clique numbers of chordal graphs
Here is another argument. A chordal graph has a *perfect elimination
ordering* $v\_1,\dots,v\_n$ of its vertices. This means that the
neighbors among $v\_1,\dots,v\_{i-1}$ of $v\_i$ form a clique $C\_i$ . See
for instance <https://en.wikipedia.org/wiki/Chordal_graph>. If $C\_i$ has
$p\_i$ vertices, then we obtain ${p\_i\choose j}$ new cliques of size
$j+1$ when we adjoin the vertex $v\_i$ to the subgraph spanned by
$v\_1,\dots,v\_{i-1}$. If $G$ is connected then $p\_i>0$ when $i>1$, so $\sum
(-1)^{j}{p\_i\choose j}=0$. Hence the sum $\sum\_{i=1}^n (-1)^{i+1}c\_i$ is
equal to the contribution from adjoining the first vertex $v\_1$,
namely, 1.


