I have no idea how to prove this I am asked to prove the equation:
$$\frac{10^n-1}9 = \underbrace{111\cdots1}_{n\space\text{digits}}$$
where $n$ is a positive integer. Can anyone help me to prove it?
 A: $\frac {10^n - 1}9 = \underbrace{1111.....111}_{n\text{ times}} \iff$
$10^n - 1 = 9\times  \underbrace{1111.....111}_{n\text{ times}}\iff$

 $10^n -1 = \underbrace{99999.....99999}_{n\text{ times}}\iff$
$10^n = \underbrace{99999.....99999}+1$

Perhaps more sophisticated is noticing $(a-1)(a^{n-1} + a^{n-2} + .....+a + 1)=$
$(a^{n} + a^{n-1} .....+a^2 + a) - (a^{n-1} + ..... + a + 1) = a^n - 1$.
Just replace $a = 10$ and ... what do you get?
A: Note that:
$$\require{cancel} \frac{10^n-1}9=\frac{\cancel{(10-1)}(10^{n-1}+10^{n-2}+\cdots +10^1+10^0)}{\cancel{9}}=\\
\underbrace{100...00}_{n}+\underbrace{100...00}_{n-1}+\cdots +\underbrace{10}_{2}+\underbrace{1}_{1}=\underbrace{111...111}_{n}.$$
A: Induction :
$\dfrac{10^1-9}{9}= 1$
Assume the formula valid for $n$.
Step $n+1:$
$\dfrac{10^{n+1}-1}{9}=$
$\dfrac{(9+1)10^n -1}{9}=$
$\dfrac{9 \cdot 10^n +(10^n-1)}{9}=$
$10^n + 1111111...(n $times$)=$
(hypothesis)
$1111111...((n+1)$ times$)).$
Note; 
$10^1 = 10$ $(1$ appears as the $2$nd digit)
$10^2 = 100 (1$ appears as the $3$rd digit)
$10^n  : 1$ appears as the $(n+1)$st digit .
A: You may simply it first. Think about it this way too. Let $n$ be 3.
$$10^3 - 1=1000 - 1 = 999$$
$$\frac{999}{9} = 111$$
Now generalize for any $n$. 
By the way, this is not a formal proof.
Added
Now, you may consider this as a kind of formal proof.
Since $10^n = 100..0$ having n zeros. Subtracting 1 from this number yeilding n nines, dividing by 9 yielding n ones. 
