Is $\frac{n}{3}! = (\frac{1}{3})^n n!$ Is $$\frac{n}{3}! = (\frac{1}{3})^n n!$$
I thought I could take all the (1/3) out of the factorial, but wolfram alpha says this is false. 
 A: false:
take: $n=6$
$\frac{n}{3}! = 2$ and 
$\frac{n}{3}! = 80/81$  
In fact the diagram below shows that equality rarely happens   

A: Take $n=3$, then LHS is $1$, but RHS is $2/9$. Hence, false.
A: The quantities are not equal; try $n=3$. As I pointed out in my comment above, you should get $1 = \frac{6}{27}$ which is clearly wrong. Factorial works nicely for whole, positive integers, but that's about the extent of it. In general it is true that $$\left(\frac{m}{n}\right)! \neq \frac{m!}{n!}$$ Factorial for fractions gets very tricky. Just read a little bit about the Gamma function to see what I mean.
A: No, compare
$$\frac{n!}{3^n}=\left(\frac{1\cdot2\cdot3}{3\cdot3\cdot3}\right)\left(\frac{4\cdot5\cdot6}{3\cdot3\cdot3}\right)\left(\frac{7\cdot8\cdot9}{3\cdot3\cdot3}\right)\cdots\left(\frac{(n-2)\cdot(n-1)\cdot n}{3\cdot3\cdot3}\right)$$
and
$$\left(\left(\frac n3\right)!\right)^3=\left(\frac{3\cdot3\cdot3}{3\cdot3\cdot3}\right)\left(\frac{6\cdot6\cdot6}{3\cdot3\cdot3}\right)\left(\frac{9\cdot9\cdot9}{3\cdot3\cdot3}\right)\cdots\left(\frac{n\cdot n\cdot n}{3\cdot3\cdot3}\right).$$
A: No, factorial function (or Gamma function) has no properties like this one which holds only for $n=0$ I guess.
A: I presume you came here from the relationship $(2n)!=2^nn!(2n-1)!!$ where the double factorial (double exclamation point) on the right says take every other term.  $9!!=9\cdot 7 \cdot 5 \cdot 3 \cdot 1$ for example.  You can prove this by looking at the terms that make up $(2n)!$. The $2^nn!$ takes care of the even ones and the $(2n-1)!!$ takes care of the odd ones.  The equivalent thing in your case (if we take !!! as multiply every third term) would be $(3n)!=3^nn!(3n-1)!!!(3n-2)!!!$.  For $n$ a multiple of $3$ you could cast this as $(\frac n3)!=(\frac 13)^nn!/(\frac{n-1}3)!!!/(\frac {n-2}3)!!!$
