User blog comment:King2218/Count to "infinity"!/@comment-3427444-20140927010624

Idea on reviving(?) this game:

Using Stirling's approximation, $$n(n-1)\cdots(m+1) \approx \sqrt{\frac{n}{m}} \frac{n^n}{m^m e^{n-m}}$$. We can start going from here.