Number times Recurring Part of Reciprocal gives 9-Repdigit

Theorem
Let a (strictly) positive integer $n$ be such that the decimal expansion of its reciprocal has a recurring part of period $d$ and no non-recurring part.

Let $m$ be the integer formed from the $d$ digits of the recurring part.

Then $m \times n$ is a $d$-digit repdigit number consisting of $9$s.

Proof
Let $x = \dfrac 1 n = \sqbrk {0. mmmm \dots}$.

Then:
 * $10^d x = \sqbrk {m.mmmm \dots}$

Therefore:

which is the $d$-digit repdigit number consisting of $9$s.