<s>[INST] How to prove divisibility of a number using the binomial expansion? I have the following problem: 
Prove that $6^n-1$ is always divisible by 5 using the binomial expansion of $(5+1)^n$. 
How can I do this? I don't know how to begin, as I don't see how the binomial expansion relates to the question. Any help would be appreciated. 
 [/INST] $$6^n-1 = (5+1)^n-1$$
$$(5+1)^n = {n \choose 0}5^n+{n \choose 1}5^{n-1}+{n \choose 2}5^{n-2}+{n \choose 3}5^{n-3}+...{n \choose n-1}5^1+{n \choose n}1$$
$$(5+1)^n-1 = {n \choose 0}5^n+{n \choose 1}5^{n-1}+{n \choose 2}5^{n-2}+{n \choose 3}5^{n-3}+...{n \choose n-1}5^1+{n \choose n}1-1$$
${n \choose n}1 = 1$ so $1$ and $-1$ cancel out.
$$\implies (5+1)^n-1 = {n \choose 0}5^n+{n \choose 1}5^{n-1}+{n \choose 2}5^{n-2}+{n \choose 3}5^{n-3}+...{n \choose n-1}5^1$$
All terms have a factor of $5$. Therefore, $6^n-1$ is divisible by $5$.
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