Dimensions of vector spaces in an exact sequence

Let us prove the version of the statement in the Wikipedia entry:

If $ 0 \to V_1 \to V_2 \cdots \to V_r\to 0$ is an exact sequence of finite dimensional vector spaces then $$\sum_{i=1}^{r} (-1)^i \dim V_i=0.$$

Let $f_0$ be the first map in the sequence, $f_i$ be the map from $V_i$ to $V_{i+1}$, etc. By the Rank Nullity theorem, we have $\dim V_i = \dim\ker f_i + \dim \operatorname{im} f_i.$ Thus the left hand side is

$$\sum_{i=1}^{r} (-1)^i \dim\ker f_i+\sum_{i=1}^{r} (-1)^i \dim\operatorname{im} f_i.$$

Now by the defining property of an exact sequence, $\operatorname{im} f_i = \ker f_{i+1}.$ Place that information into one of the sums, and the two sums then cancel out.

Note that in order for the series in question to converge, the sequence must be of the form in this answer, perhaps with extra $0$'s.