<s>[INST] Prove $\binom{n}{2}+\binom{n}{3}\cdot7+\binom{n}{4}\cdot7^2+....+7^{n-2}\in \mathbb{Z}$ for all $n\geq 2$ 
If $X=\{8^n-7n-1:n\in \mathbb{N}\}$ and $Y=\{49(n-1):n\in \mathbb{N}\}$ then show that $X⊂Y$

While solving the problem
$$8^n-7n-1=(1+7)^n-7n-1\\
=49\bigg[\binom{n}{2}+\binom{n}{3}\cdot7+\binom{n}{4}\cdot7^2+...+\binom{n}{n-1}.7^{n-3}+7^{n-2}\bigg]\\
$$
How do I know that the expression inside the bracket is an integer for all $n\geq 2$ ?
 [/INST] Another way to expand using
$$a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+a^2b^{n-3}+ab^{n-2}+b^{n-1}\right) \tag{1}$$
is
$$8^n-7n-1=8^n-1-7n=\color{red}{(8-1)}\left(8^{n-1}+8^{n-2}+...+8+1\right)-\color{red}{7}n=\\
7\left(8^{n-1}+8^{n-2}+...+8+1-\color{blue}{n}\right)=...$$
there are $\color{blue}{n}$ terms $8^{n-1},8^{n-2},...,8,1$, thus
$$...=7\left(8^{n-1}-\color{blue}{1}+8^{n-2}-\color{blue}{1}+...+8-\color{blue}{1}+1-\color{blue}{1}\right)=\\
7\left[\color{red}{(8-1)}\left(8^{n-2}+8^{n-3}+...+8+1\right)+\color{red}{(8-1)}\left(8^{n-3}+8^{n-4}+...+8+1\right)+...+\color{red}{7}\right]=\\
49\left[\left(8^{n-2}+8^{n-3}+...+8+1\right)+\left(8^{n-3}+8^{n-4}+...+8+1\right)+...+1\right]$$
Your version is a lot shorter and, as per the comments, correct one (binomial coefficients are integers). Sometimes it's good to have more versions to validate the answer.
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