Let $n$ be a positive integer such that $\gcd (n, 10) = 1$. Prove that $n \mid 99 \cdots 99$ where there are $\phi (n)$ number of 9’s in $99 \cdots 99$.
Since $\gcd(n,10)=1$, we can use Euler's Theorem
$$ 10^{\phi(n)} \equiv 1 \pmod {n} $$
This immediately gives us
$$ 10^{\phi(n)} -1 \equiv 0 \pmod {n} $$
which means
$$ \begin{align} n & \mid 10^{\phi(n)} -1 \\ \\ n & \mid 99 \cdots 99 \end {align}$$
The last step uses $10^{\phi(n)} - 1 = 99 \cdots 99$, a decimal number with $\phi(n)$ nines.