Minimum value of a big function For a positive integer $n≥2$, let
$$f(n)=\frac{(log_32)(log_33)(log_34)⋯(log_3n)}{9^n}$$
Let $m$ denote the minimum value of $f(n)$, and let $N$ be the sum of all values of $n$ for which $f(n)=m$. Find $N$. I need help simplifying the function. 

Note that for $n>2$
$$f(n)=f(n-1)\cdot\dfrac{\log_3 n}{9}.$$
For $n<3^9$ $\quad f(n)<f(n-1)$.
For $n=3^9$ $\quad f(n)=f(n-1)\cdot \dfrac{\log_3 3^9}{9}=f(n-1)$.
For $n>3^9$ $\quad f(n)>f(n-1)$.
So there are two minimal values: $f(3^9-1)$ and $f(3^9)$.
$N = (3^9-1)+3^9$.
