Comparing sets $\{8^n-7n-1: n \in \Bbb{N}\}$ and $\{49(n-1): n \in \Bbb{N}\}$ 
If
$$\begin{align}
X&=\{8^n-7n-1: n \in \Bbb{N}\} \\
Y&=\{49(n-1): n \in \Bbb{N}\}
\end{align}$$
then, 
a) $X\subset Y \qquad$ b) $Y\subset X\qquad$ c) $X=Y\qquad$d) none of these

I know this question can be solved by taking $X=8^n-7n-1$ and splitting $8^n$ into $(7+1)^n$ and then apply binomial theorem as follows:
Given,
$$\begin{align}X &=8^n−7n−1 \\[4pt]
&=(1+7)^n−7n−1 \\[4pt]
&=1+7n+\frac{n(n-1)}{2}+\cdots+7^n-7n-1 \\[4pt]
&=\frac{n(n-1)}{2}7^2+\cdots+7^n \\[4pt]
&=49\left[\frac{n(n-1)}{2}+\cdots+7^{n-2}\right]
\end{align}$$
Hence, the set $X$ will be some specific multiples of $49$.

$Y=49(n-1)$. Hence, the set $Y$ will be all multiples of $49$. So, it will contain the elements of $X$ too.
So, $$X\subset Y$$

But is there any alternate/simple method to solve this question without using binomial theorem?

 A: Here's an alternative method, using dxiv's suggestion to invoke the difference-of-powers identity:
Let $n\in\mathbb N.$
$$8^n-7n-1\\=(8-1)(8^0+8^1+\ldots+8^{n-1})-7n$$$$=7(8^0+8^1+\ldots+8^{n-1}-n).\tag1$$
Working $\mod7,$ $$8^0+8^1+\ldots+8^{n-1}-n\\\equiv n\times1-n$$$$=0.\tag2$$
From $(1)$ and $(2),$ and since $8^n-7n-1$ is increasing with $n,$ and equals $0, 49$ and $490$ when $n$ equals $1,2$ and $3,$ respectively, $8^n-7n-1$ is some nonnegative multiple of $49$ but is not $2\times49.$
Hence, for some proper subset $A$ of $\mathbb N,$ $$X=\{8^n-7n-1\mid n \in \mathbb N\}\\=\{49(k-1)\mid k \in A\}.$$
But $$\{49(k-1)\mid k \in A\}\\\subset\{49(n-1)\mid n \in \mathbb N\}=Y.$$
Therefore $$X\subset Y.$$
A: Induction:
$n=0$: $49|8^0-7\cdot 0-1$
The difference between $n$ and $n+1$ is $8^{n+1}-8^n-7=7(8^n-1)$ which is divisible by $49$, so if $49$ divides the $n$th term, it divides the following one.
A: Putting $n=1,2,3,4,5,…$,
we obtain: $X=\{0,49,490,4067,32732,…\}$.
Similarly, putting $n=1,2,3,…,11,…,84,…,669$,
we have: $Y=\{0,49,98,...,490,…,4067,…,32732,…\}$.
$X=\left\{8^{n}-7 n-1 \mid n \in N\right\}=\{0,49,490, \ldots\}$
$Y=\{49 n-49 \mid n \in N\}=\{0,49,98,147, \ldots, 490, \ldots\}$
Clearly, every element of $X$ is in $Y$ but every element of $Y$ is not in $X$.
$\therefore \quad X \subset Y$
A: $8^n - 7n - 1 = (8^n - 1) - 7n = (8-1)(8^{n-1} + ..... + 8 + 1) - 7n = 7[(8^{n-1} + .... + 8 + 1) - n]$.
So $7|8^n - 7n - 1$.
And $8^n \equiv 1^n \equiv 1 \pmod 7$ so $(8^{n-1} + .... + 8 + 1) - n\equiv (1+1+1+....+1) -n \equiv n - n \equiv 8 \pmod 7$
So $7|(8^{n-1} + .... + 8 + 1) - n$ and $49|8^n - 7n - 1=7[(8^{n-1} + .... + 8 + 1) - n]$.
As $Y$ is the set of all non negative multiple of $49$ and $8^n-7n - 1$ is a non-negative multiple of $49$ we know $X\subset Y$.
But as $n$ gets large $8^n -7n -1$ will increase by larger amounts than $49$ and not all multiples of $49$ can possibly be covered by $8^n-7n-1$.
Indeed for $n = 2$ we have $8^2 - 7\cdot 2 -1 = 49$ but for $n =3$ we have $8^2 -3\cdot 7 -1 = 490$ and we never have any multiples of $49$ between $49=49\cdot 1$ and $490 = 49\cdot 10$.
And furthermore  If $A_k = 8^k - 7k - 1$ then $D=A_{k+1} - A_k = 7\cdot 8^{k+1} - 7$. and $k> 1$ then $D > 49$ and $A_n$ is increasing and by more than $49$.
So $Y\not \subset X$.
