Recursive factorial algorithm
If you plug in the result in what is returned by the recursion, you get:
$$\frac{n!}{(n - \lfloor \frac{m}{2} \rfloor)!}
\frac{(n - \lfloor \frac{m}{2} \rfloor)!}{(n - \lfloor \frac{m}{2} \rfloor - \lceil \frac{m}{2} \rceil)!}
= \frac{n!}{(n - \lfloor \frac{m}{2} \rfloor - \lceil \frac{m}{2} \rceil)!}$$

You can see the right side is equal to $\frac{n!}{(n - m)!}$ because of the identity $m = \lfloor \frac{m}{2} \rfloor + \lceil \frac{m}{2} \rceil$ (if you are in doubt about this: check the cases $m$ even and $m$ odd separately).

For correctness, the only thing left to check then is the base cases $m=0$ and $m=1$ are correct (which is a trivial observation), that Falling is not invoked with incorrect parameters (this is the case if $m \le n$ holds) and that it terminates (also the case because the invoked instances have strictly smaller parameters).