There are several (and sometimes controversial) definitions of a cyclic number. I take the easiest to deal with:
n-digit number is cyclic if its products by $2,\,3\dots (n-1)$ are cyclic permutations of the original number
The best-known cyclic number is $142857=\frac{10^6-1}{7}$ $$ \begin{array}\\ 142857 \times 2 = 285714\; &\; 142857 \times 3 = 428571\; &\; 142857 \times 4 = 571428 \\ 142857 \times 5 = 714285\; &\; 142857 \times 6 = 857142 \end{array} $$
Another one is $0588235294117647 = \frac{10^{16}-1}{17}$ (Unfortunately, zero at start cannot be omitted)
I can prove that if $p$ is a full-repetend prime (i.e. the repetend of $\frac{1}{p}$ consists of $p-1$ digits) then $\frac{10^{p-1}-1}{p}$ (Fermat quotient) is a cyclic number. In other words, if $p$ is a full-repetend prime, then the repetend of $\frac{1}{p}$ is a cycling number.
Wikipedia article claims also the opposite: Any cyclic number can be represented as a Fermat quotient of a full-repetend prime. (Any cycling number is a repetend of $\frac{1}{p}$ for some full-repetend $p$). Quite obviously it's sufficient to demonstrate that $a(n+1)=10^n-1$, for $n$-or-less-digit cycling number $a$. Any idea of proving that?