Solving recurrence relation: Product form Please help in finding the solution of this recursion. 
$$f(n)=\frac{f(n-1) \cdot f(n-2)}{n},$$
where $ f(1)=1$ and $f(2)=2$.

Take some large n,
$$
\begin{align}
f(n) &= \frac{f(n-1) f(n-2)}{n} \\
&= {f(n-2)^2 f(n-3) \over n (n-1)} \\
&= {f(n-3)^3 f(n-4)^2 \over n (n-1) (n-2)^2} \\
&= \frac{f(n-4)^5 f(n-5)^3}{n (n-1) (n-2)^2 (n-3)^3} \\
&= \dots \\
&=\frac{f\left(n - (n-2) \right)^{F_{n-1}} f\left(n - (n-1) \right)^{F_{n-2}}}{\prod^{n-3}_{i=0} (n-i)^{F_{i+1}}} \\
&=\frac{2^{F_{n-1}}}{\prod^{n-3}_{i=0} (n-i)^{F_{i+1}}}
\end{align}
$$
in each step until the "..." I just plugged the formula into the left factor of the numerator until I saw a pattern. The symbol $F_n$ denotes the Fibonacci sequence, which follows $F_n = F_{n-1} + F_{n-2}$ and $F(0) = 0$ and $F(1) = 1$. Sorry this is not a formal proof, but you can easily prove it by induction by showing that, if it's true for some large n then it must be true for n+1, and then simply demonstrating that it's true for 3.
