Question: > 
> The series $(a\_n)\_{n\in\mathbb{N}}$ is given through 
>  $$a\_1=1,\quad a\_2=\frac{1}{2},\quad a\_{n+2}=a\_na\_{n+1} \quad\text{ for } n\geq1.$$
>  I want to show that $$a\_n = 2^{-f\_{n-1}}$$ whereas $\,f\_n$ is the $n$-th Fibonacci number.
> 
> 
> 

I did a proof with induction, but now I came up with this. Out of the initial figures, the definition of $f\_n$, and the exponent and logarithm rules I think it is

\begin{align}
 &\quad a\_{n} = 2^{\,\log\_2 a\_n} = 2^{\,\log\_2 (a\_{n-2} \,\cdot\, a\_{n-1})}\\
 \Longleftrightarrow &\quad a\_n = 2^{\,\log\_2 (a\_{n-2}) \,+\, \log\_2 (a\_{n-1})}\\
 \Longleftrightarrow &\quad \log\_2 a\_n = \log\_2 2^{\,\log\_2 (a\_{n-2}) \,+\, \log\_2 (a\_{n-1})}\\
 \Longleftrightarrow &\quad \log\_2 a\_n = \log\_2 a\_{n-2} \,+\, \log\_2 a\_{n-1}\\
 \Longleftrightarrow &\quad (-\log\_2 a\_n) = (-\log\_2 a\_{n-2}) + (-\log\_2 a\_{n-1}) \quad(\*)
\end{align}

and with the initial figures $-\log\_2 a\_1 = 0$ and $-\log\_2 a\_2 = 1$ the negative expononents form the Fibonacci sequence $2^{-f\_{n-1}}$ with $(\*)$.

Is that valid? Thanks!
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Answer: 
I think that you should at least mention that $a\_n\gt 0$ for every $n$.

Note that taking the logarithm of $a\_{n+2}=a\_na\_{n+1}$ directly is simpler :
$$\log\_2 a\_{n+2}=\log\_2 a\_n+\log\_2 a\_{n+1}$$
Your proof is fine. However, I'd suggest another version. Initialization: $a\_1 = 2^{-f\_0}$ (assuming $f\_0 =0$), $a\_2 = 2^{-f\_1}$ (assuming $f\_0 =1$). It happens that $a\_3 = a\_1\*a\_2 = 2^{-f\_0-f\_1} = 2^{-f\_2}$ . Now suppose the property is valid up to index $k\ge 2$. Then $a\_{k+1} = a\_{k}a\_{k-1} = 2^{-f\_{k-1}}2^{-f\_{k-2}} = 2^{-f\_{k-1}-f\_{k-2}} = 2^{-f\_{k}}$. So it is valid for index $k+1$ as well. 

Thus, you do not have to play with $\log$s, which requires positivity.