Show membership for all naturals I am pretty confused by the approach to consider (of course, except mathematical induction; that seems not correct for considering all cases possible) question as below. Mathl. induction fails, and I am not sure of using it.
Show that the numbers generated by the formula are all lying in the set of 
naturals, for n being a natural number:
$\frac{(3n)!}{3^n} \in N$

$$\prod_{r=1}^{3n}r=\prod_{m=1}^n(3m-2)(3m-1)(3m)=3^n\prod_{m=1}^nm(3m-2)(3m-1)$$
