Cyclic numbers are characterized by the reciprocals of full reptend primes? The number $142,857$ is widely known as a cyclic number, meaning consecutive multiples are cyclic permutations, i.e.
$1 × 142,857 = 142,857$
$2 × 142,857 = 285,714$
$3 × 142,857 = 428,571$
and so on. 
142857 is the repeating unit of $\frac{1}{7} = 0.\overline{142857}$ and in fact, every prime for which $10$ is a primitive root will generate a cyclic number (if we allow $0$ as a first digit, for example $0588235294117647$ which is the repeating unit of $\frac{1}{17}$). These primes are called full reptend primes.
From what I've read, it seems that there is a bijection between full reptend primes and cyclic numbers: a number is cyclic if and only if it is the repeating unit for the reciprocal of a full reptend prime.
I was able to find a proof for the if part. I was wondering if anyone can provide a proof for the only if part.
Edit: It's been a while since I posed this question and I still haven't found a proof. I've added a bounty in hopes of prompting some interest.

The only if part is: if some number $n$ in base $b$ of length $d=\lfloor{\log_b n}\rfloor+1$ cycles for $2n$ through $(d-1)n$ then $n/(b^d-1)=1/k$, $k$ prime and not a factor of $b$.
(This definition excludes the solutions with leading zeros.)
I think the approach is to show $dn=n+(d-1)n$ must equal $b^d-1$ and hopefully the rest follows...
Let $$  n=n_1 n_2\dots n_d $$
then $$ n/(b^d-1)=0.\overline{n_1 n_2\dots n_d} $$
so $$ bn/(b^d-1)=n_1.\overline{n_2\dots n_d n_1}=n_1+m_1n/(b^d-1) \text{ for some $m_1$ between $2$ and $d-1$} $$
because $n$ is cyclic.
$$ \therefore b=n_1(b^d-1)/n+m_1 $$
and similarly 
$$ b^i=n_1n_2\dots n_i(b^d-1)/n+m_i, 2 \le  m_i \le d-1, m_d=1, 1\le i \lt d, m_i \text{ distinct } $$
So all of $n_1n_2\dots n_i(b^d-1)/n$ must be integers.
Specifically
$$ n_1n_2\dots n_i(b^d-1)/n = b^i-m_i $$
$$ \therefore (b^d-1)/n = (b^i-m_i) / n_1n_2\dots n_i $$
and reciprocating...
$$ n/(b^d-1) =  n_1n_2\dots n_i / (b^i-m_i), 2 \le  m_i \le d-1, m_d=1, 1\le i \lt d, m_i \text{ distinct } $$
etc... Now just to show $ n/(b^d-1) = 1/k, k $ prime and not a factor of $ b $ :-) ...
