Binomial Formula evaluation for the problem $9 \mid 10^k - 1$ for $k \in \mathbb{N}$

Question

Prove using binomial formula $9 \mid 10^k - 1$ for $k \in \mathbb{N}$

I am aware of similar answer to this question here and here however my query is about the manipulation on the binomial formula in my proof.

My attempt at the proof :

We notice that $$10^k - 1 = (9 + 1)^k - 1$$ $$=\left(\sum_{i = 0}^{k} {k \choose i} 9^{k-i} \right) - 1$$

How do I further remove the -1 to show that this indeed is divisible by 9 ?

Answer 1

You can't really remove the $-1$, but don't need to: one of the terms in the sum is $1$, cancelling with the $-1$. The rest are divisible by $9$.

Answer 2

Hint

$$\left(\sum_{i = 0}^{k} {k \choose i} 9^{k-i} \right) - 1=$$ $$=\sum_{i = 0}^{k-1} {k \choose i} 9^{k-i}+ {k \choose k} 9^{k-k} - 1$$

How much is ${k \choose k} 9^{k-k}$?