Let $K$ be an oriented (an ordering of the vertices) simplicial complex of dimension $s$ (meaning that $K$ does not contain an $n$-simplex for $n>s$). If $r_n$ denotes the number of $n$-simplices in $K$, we define the Euler characteristic of $K$ as:
$$ \sum_{n=0}^s (-1)^n r_n $$
I want to prove that this is equal to:
$$ \sum_{n=0}^s (-1)^n rank( H_n(K) ) $$
Where $H_n(K)$ is the simplicial homology of $K$.
To expand on the comment of @Pedro, here is the theorem that does what you want (and it is the theorem to which Hatcher is referring in his "purely algebraic" comment; you will often see this theorem stated separately in other algebraic topology books, such as Spanier).
Suppose you have a finite chain complex $$0=C_{N+1} \to C_N \to C_{N-1} \to \cdots \to C_1 \to C_0 \to C_{-1}=0 $$ such that each $C_n$ is a free abelian group of finite rank. Denote the homology groups of this chain complex to be $$H_n = \text{kernel}(C_n \to C_{n-1}) / \text{image}(C_{n+1} \to C_n), \quad \text{$0 \le n \le N$} $$ Each group $H_n$ is a finitely generated abelian group, hence is a direct sum of a free abelian group of finite rank denoted $\text{rank}(H_n)$ plus a finite abelian group that is not relevent to the present discussion.
Then we have $$\sum_{i=0}^n (-1)^n \, \text{rank}(C_n) = \sum_{i=0}^n (-1)^n \, \text{rank}(H_n) $$ In the case of a finite simplicial complex and its associated simplicial chain complex, $\text{rank}(C_n)$ is equal to the number of $n$-simplices, and $H_n$ is the simplicial homology.