Prove that the number $ \underbrace{11 \ldots 1}_{(p-1) \mathrm{l}^{\prime} \mathrm{s}} $ is divisible by $p$ Question -

Example 1.30. Let $p \geq 7$ be a prime. Prove that the number
$
\underbrace{11 \ldots 1}_{(p-1) \mathrm{l}^{\prime} \mathrm{s}}
$
is divisible by $p$

Proof: We have
$
\underbrace{11 \ldots 1}_{p-1^{\prime} \mathrm{s}}=\frac{10^{p-1}-1}{9}
$
and the conclusion follows from Fermat's little theorem. 

But by FLT we get ${10^{p-1}-1} \equiv 0 \pmod{p}$ how we will get that $\frac{10^{p-1}-1}{9} \equiv 0 \pmod{p}$?
Well, simply
$$9 | (10^{p-1}-1) \ \ \text{and} \ \ p|(10^{p-1}-1)$$
So,
$$10^{p-1}-1=9pk$$
where $\gcd(9,p)=\gcd(9,k)=\gcd(p,k)=1$ for some $k \in \mathbb{Z^{+}}$. So
$$\frac{10^{p-1}-1}{9}=pk \implies p|\left(\frac{10^{p-1}-1}{9}\right) \ \Box.$$

Note: $9 | (10^{p-1}-1)$ because of the factoring formula 
$$x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+ \dots + xy^{n-2}+y^{n-1})$$
Plugging in $x=10$, $y=1$ and $n=p-1$ gives the conclusion.
Since $p$ is not divisible by $3$, modulo $p$ we have that division by $9$ corresponds to multiplication by some constant. Which constant will depend on $p$, but for instance, for $p=7$ we have $\frac19\equiv 4$, and for $p=11$ we have $\frac19\equiv 5$.

This gives us $10^{p-1}-1\equiv 0\implies \frac{10^{p-1}-1}{9}\equiv 0$.