$\pi$ (...) something!

By the binomial theorem,

$\pi(n) = {n \choose 0} {n \choose 1} \ldots {n \choose n} = \frac{n!^{n+1}}{0!^{2} 1!^{2} \ldots n!^{2}}$

From this, we then know

$\frac{\pi(n+1)}{\pi(n)} = \frac{(n+1)!^{n+2}}{n!^{n+1}} \cdot \frac{0!^{2} 1!^{2} \ldots n!^{2}}{0!^{2} 1!^{2} \ldots (n+1)!^{2}} = \frac{(n+1)^{n}}{n!}$

Subsituting in $n = 2001$ gives our desired answer of $\boxed{4002}$ .

Please change question to find the lowest value of m+n as (2002,2002) and (2001,2001) are both valid solutions.

2002^2002/2002!=(2002^2001) 2002/((2001!) 2002)=2002^2001/2001!