The Euler–Poincaré theorem

Fix $n$ and consider the boundary map $\partial_n : C_n \to C_{n-1}$ as a linear map of finite-dimensional vector spaces (working over $\mathbb{Q}$ for simplicity). By the rank-nullity theorem, $\operatorname{rank} C_n = \dim \ker \partial_n + \operatorname{rank} \partial_n$, where $\operatorname{rank}\partial_n := \dim \operatorname{im} \partial_n$.

Also, $\dim H_n(X) = \dim \ker \partial_n - \dim \operatorname{im} \partial_{n+1}$ directly from the quotient definition $H_n = \ker \partial_n / \operatorname{im}\partial_{n+1}$, so $\dim \ker \partial_n = \beta_n + \operatorname{rank}\partial_{n+1}$.

Substituting, $\operatorname{rank} C_n = \beta_n + \operatorname{rank}\partial_{n+1} + \operatorname{rank}\partial_n$. Now form the alternating sum over all $n$ from $0$ to the top dimension $N$: $\sum_n (-1)^n \operatorname{rank} C_n = \sum_n (-1)^n \beta_n + \sum_n (-1)^n \operatorname{rank}\partial_{n+1} + \sum_n (-1)^n \operatorname{rank}\partial_n$.

In the last two sums, the term $\operatorname{rank}\partial_k$ appears once from the $\operatorname{rank}\partial_n$ sum (at $n=k$, with sign $(-1)^k$) and once from the $\operatorname{rank}\partial_{n+1}$ sum (at $n=k-1$, with sign $(-1)^{k-1}$); these two signs are opposite, so every $\operatorname{rank}\partial_k$ cancels exactly (telescoping), except the boundary terms $\partial_0 = 0$ and $\partial_{N+1}=0$ which contribute nothing anyway.

What remains is $\sum_n (-1)^n \operatorname{rank} C_n = \sum_n (-1)^n \beta_n$, and since the left side is $\chi(X)$ by definition, this is exactly $\chi(X) = \sum_{n\ge 0} (-1)^n \operatorname{rank} C_n = \sum_{n\ge 0} (-1)^n \beta_n(X)$.