It's not possible to count all the small triangles. The best one can do is to count

of them, and this is why. (Press the "Show hint" button.)

Yes, it is not difficult to construct a chain of adjacent triangles of length n² - n + 1. Think of the partition of the big triangle into the smaller once as being composed of (horizontal) layers. The lowest layer contains 2n-1 triangle, the next one 2n-3, ..., and the very last contains 1 (top) triangle.

In each layer, all triangles but either the leftmost or the rightmost form a parallelogram which is a chain of triangles with the exit and entry points at the opposite corners. Thus in each layer there are two possible parallelograms. The parallelograms can be now stacked on top of each other continuously so that the exit edge of a layer coincides with the entry edge of the next layer above. In all, the chain will include

triangles. Exactly as in (1).

Nathan Bowler offers a proof by induction:

Base case: For n = 1, the triangle itself is a chain of length 1 = 1² - 1 + 1.

Induction step: Pick a side adjacent to the given corner. The triangles with at least one vertex on that side form a chain of length 2n - 1. Take all the elements of that chain except the final one. The triangles not in that chain form a triangle of side n-1, of which you are now adjacent to the corner. By the induction hypothesis, then, we may extend the chain by a further (n-1)² - (n-1) + 1 triangles giving a total length of n² - 2n + 1 - n + 1 + 1 + 2n - 2 = n² - n + 1.

Of course, this proof does not give an explicit construction, but it suggests two reasonably pleasant ones: a zigzag and a spiral.