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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 ?

Krosin
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    Every term in the summation but the last is a multiple of $9$ and the last term in the summation is equal to $1$ itself and cancels with the $-1$ outside of the summation. – JMoravitz May 07 '21 at 15:08
  • Oh I see, Thanks! – Krosin May 07 '21 at 15:09
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    @JMoravitz pls write this comment as answer so that OP can accept your answer. –  May 07 '21 at 15:41

2 Answers2

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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 ${k \choose k} 9^{k-k}$?

Arnaldo
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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 $9$.