Solution:

We shall show that $f(n) = (n-1)!$.

Label the horizontal line segments in the triangle $l_{1}, l_{2}, \ldots$ as in the diagram below. Since the path goes from the top triangle to a triangle in the bottom row and never travels up, the path must cross each of $l_{1}, l_{2}, \ldots, l_{n-1}$ exactly once. The diagonal lines in the triangle divide $l_{k}$ into $k$ unit line segments and the path must cross exactly one of these $k$ segments for each $k$. (In the diagram below, these line segments have been highlighted.) The path is completely determined by the set of $n-1$ line segments which are crossed. So as the path moves from the $k$th row to the $(k+1)$st row, there are $k$ possible line segments where the path could cross $l_{k}$. Since there are $1 \cdot 2 \cdot 3 \cdots (n-1) = (n-1)!$ ways that the path could cross the $n-1$ horizontal lines, and each one corresponds to a unique path, we get $f(n) = (n-1)!$.

Therefore $f(2005) = (2004)!$.