# The period of decimal expansion of $1/m$

I am trying to prove that for a composite, positive integer $m$ such that $2 \nmid m$ and $5 \nmid m$, the period of the decimal expansion of $1/m$ is equal to $\text{ord}_{m}(10)$, where $\text{ord}_{m}(10)$ denotes the multiplicative order of $10$ mod $m$.

So let $k \in \mathbb{N}$ be the period of $1/m$ so that we have $1/m = 0.\overline{a_{1}a_{2}\cdots a_{k}}$. Then it follows that $10^k \cdot 1/m = a_{1}a_{2} \cdots a_{k}. \overline{a_{1} a_{2} \cdots a_{k}}$, and so $$10^k/m - 1/m = a_{1}a_{2} \cdots a_{k}. \overline{a_{1} a_{2} \cdots a_{k}} - 0.\overline{a_{1}a_{2}\cdots a_{k}} \in \mathbb{Z}.$$

If we let $a_{1}a_{2} \cdots a_{k} = n$, then we have that $$\frac{1}{m}(10^k -1) = n \Leftrightarrow 10^k -1 = nm,$$ and so $m \mid 10^k -1$ or equivalently that $10^k \equiv 1 \mod m$. At this point I can conclude that $\text{ord}_{m}(10) \leq k$, but how do I see that $k \leq \text{ord}_{m}(10)$?

EDIT: So I think I nailed down why $k \leq \text{ord}_{m}(10)$. So let $t = \text{ord}_{m}(10)$, then we have that $10^t \equiv 1 \mod m$ and so $m \mid 10^t - 1$ or, equivalently, $10^t - 1 = mr$ for some $r \in \mathbb{Z}$. Dividing both sides of the equation above by $m$, we have that $$\frac{1}{m} (10^t - 1) = r,$$ or, equivalently, $$\frac{10^t}{m} - \frac{1}{m} = r.$$ Now if $t$ was not a multiple of $k$, then $10^t/m - 1/m$ would not be an integer. However since $10^t/m - 1/m$ is clearly an integer, it must be the case that $t$ is a multiple of $k$ and so $k \leq t$. Therefore the multiplicative order of 10 mod m is equal to the period of $1/m$.

Is the sentence "Now if $t$ was not a multiple of $k$, then $10^t/m - 1/m$ would not be an integer" clear enough, or should I explain further?