A proof about period length In this page: http://mathworld.wolfram.com/FermatPrime.html
I found the following result:
$2^{2^{k}}+1$ is a Fermat prime if and only if the period length of $1/(2^{2^{k}}+1)$ is $2^{2^{k}}$.
I am asking where I can find this proof.
 A: First of all : The statement is not true for $k=0$ and $k=1$. But for $k>1$ there is a proof which however needs some advanced number theoretical tools.
The period length of $1/N$ for some positive integer $N$ coprime to $10$ is equal to $$ord_{10}(N)$$ which is equal to the smallest positive integer $u$ such that $$10^u\equiv 1\mod N$$ holds.
A Fermat number (let us call it $P$) larger than $5$ is coprime to $10$. What we have to show is that we have 

$ord_{10}(P)=P-1$ if and only if $P$ is prime.

First suppose that $P$ is prime. Then, we have $$10^{P-1}\equiv 1\mod P$$ because of Fermat's little theorem and because of Euler's criterion we have 
$$10^{(P-1)/2}\equiv \left(\frac{10}{P}\right)\mod P$$ 
where () denotes the Legendre-symbol.
Since $k>1$ , $P$ is of the form $8k+1$ , hence $2$ is a quadratic residue mod P.
$P$ must also be of the form $5k+2$, hence $5$ is not a quadratic residue mod P.
Hence $10$ is not a quadratic residue mod P and we get $ord_{10}(P)=P-1$
If we have $ord_{10}(P)=P-1$ , we can immediately conclude that $P$ is prime (this is not only true for Fermat numbers). This finishes the desired proof.
