Why is $10^k - 1$ divisible by $9$? I know it is obvious that $10^k-1$ will always be divisible by $9$ for some integer $k$, but I am curious how to actually prove this.
$$10^k - 1 \equiv 0 \bmod 9$$
$$10^k \equiv 1 \bmod 9$$
... and I have no idea where to go from here.
 A: It's just that $10$ is $1$ modulo $9$, and modulo being compatible with multiplication. So $10^k \equiv 1^k \equiv 1 \mod 9$. 
More generally this shows that $(a+1)^k \equiv 1 \mod a$. Indeed, writing it this way suggests another proof: 
Expanding $(a+1)^k$ using the binomial theorem one see it is of the form $1+ a N$ for some integer $N$.  
A: Since you've tagged modular arithmetic: $$\begin{align*}10 \equiv 1 \pmod{9}   \implies 10^k &\equiv 1^k \pmod{9} \\   10^k &\equiv 1 \pmod{9} \\   10^k  -1 &\equiv 0\pmod{9}\end{align*}$$
A: The question seems to be:
Why is $\underbrace{ 99\dots 99}_n$ a multiple of $9$?
It is clear, since it is $\underbrace{ 11\dots 11}_n\times 9$
A: Other Way
$$10^k-1=(10-1)(10^{k-1}+10^{k-2}+\cdots+1)$$
A: When a question involves specific number (in this case $10$) one has to use special property of that number.
The special property of 10  relevant to our interest here is that  it is one more than 9. Now we can use binomial theorem to expand
$10^k =(9+1)^k$. In the binomial expansion on RHS except the final term $1^k$ all other terms are multiples of powers of $9$. So moving that exceptional term to LHS  we get  $10^k-1$ is a multiple of $9$.
A: By induction: When $k = 1$, $10^k - 1 = 9$ is divisible by $9$.
Now suppose $k \geq 2$, and $10^{k-1}-1$ is divisible by $9$.  By induction, we can write $10^{k-1} - 1 = 9a$ for some integer $a$.  Then $$10^k - 10 = 90a$$ Add $9$ to both sides: $$10^k - 1 = 90a + 9 = 9(10a+1)$$ so $10^k - 1$ is divisible by $9$.
