If a category $\mathcal{C}$ has all binary equalizers, does it necessarily have all finite equalizers?
If we take $\mathcal{C}=Sets$ then we can always take the intersection of two equalizers with its inclusion mapping and at worst end up with the empty set and its trivial inclusion mapping, and more generally any category with an initial object has all equalizers trivially, but it isn't clear to me that binary equalizers in a vacuum yield all finite equalizers.
Let's show how to get $(k+1)$-fold equalisers from binary equalisers and $k$-fold equalisers. Let $f_1,\ldots,f_{k+1}$ be parallel arrows. Let $g$ be an equaliser of $f_1,\ldots,f_k$ and $h$ an equaliser of $f_1\circ g$ and $f_{k+1}\circ g$. Then $g\circ h$ is an equaliser of $f_1,\ldots,f_{k+1}$.