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Consider the sequence: $$ x_1=2; x_{n+1}=2^{x_n}-1 \text{ for } n\ge 2$$ Prove that $x_n$ divides $2^{x_n}-2$.

I tried writing $2^{x_n}-2=2^{2^{x_n-1}-1}-2$ and so on, but that does not help much. Any hints?

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    This question lacks the sort of context that we look for on this site. Well-composed questions include more than just a problem: they include discussion of why the problem is of interest, and the general source of the problem. In this case, there is a clear appearance that the problem came from somewhere, but nothing is mentioned in the question. Knowing the background also helps others write answers at an appropriate level. Mentioning a basic idea you've tried is not as informative as describing the background of the problem. You can edit the post to include additional information. – 2017-01-08
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    @CarlMummert I admit, I've been aware of the flaws of my question you mention, I'll do my best to improve the question appropriately. – 2017-01-08

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The proof is indeed by induction. For $n = 1$, it is true. Now suppose that $x_n \mid 2^{x_n} - 2$ holds for $n$. Write $$2^{x_{n+1}} - 2 = 2^{2^{x_n} - 1} - 2 = 2\left( 2^{2^{x_n} - 2} - 1\right).$$ Since $x_{n+1}$ is odd, we have to show that it divides $2^{2^{x_n} - 2} - 1$. This brings to mind the following factorization for any $z$: $$z^a - 1 = (z-1)(z^{a-1} + z^{a-2} + \cdots + 1).$$

In particular,

if $a \mid b$, then $y^a - 1 \mid y^b - 1$. (Take $b = na$, $z = y^n$).

Then apply the inductive hypothesis that $x_n \mid 2^{x_n} - 2$.

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Suppose for induction $\ x_n\mid 2^{\large x_n}\!-2.\,$ Then $\ \color{#0a0}{2^{\large x_n}}\! = \color{#0a0}{2\!+\!kx_n}\ $ therefore

${\rm mod}\,\ x_{n+1}=\color{#c00}{2^{\large x_n}\!-1}\!:\ \ 2^{\large x_{n+1}}= 2^{\large \color{#0a0}{2^{\Large x_n}}-1} = 2^{\large\color{#0a0}{ 1+kx_n }} = 2(\color{#c00}{2^{\large x_n}})^{\large k}\equiv 2\ $ by $\,\color{#c00}{2^{\large x_n}\equiv 1}$

thus $\ x_{n+1}\mid 2^{\large x_{n+1}}\!-2,\ $ i.e. $\ P(n)\,\Rightarrow\, P(n\!+\!1),\,$ so the induction step is true.