TITLE: Show $\underbrace{{111\cdots}1}_{{\small{p-1} \ 1's}}$ is divisible by $p$
QUESTION [5 upvotes]: What is the shortest proof to show $\underbrace{{111\cdots}1}_{{\small{p-1} \ 1's}}$ is divisible by $p$

REPLY [10 votes]: $$\underbrace{{111\cdots}1}_{{\small{p-1} \ 1's}} = \frac{10^{p-\small{1}}-1}{9} \equiv 0 \pmod{p}$$
because by Fermats little theorem, since gcd$(10,p)=1, 10^{p-1} \equiv 1 \pmod{p}$
NOTE: Assuming $p > 5$. The result is not true for $p=2, 3$ or $5$. In all the excitement to show, I missed important point. 
For $p=2$ obviously $1$ is not divisible by $2$ - does not hold, and 
$p=3$, $11$ is not divisible by $3$ and finally for $p=5$ gcd$(5,10)=2$ and therefore not true as well ($1111$ is not divisible by $5$).