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I tried to play around with $2^{2^{n + 1}} - 1$, such as splitting it to $2^{2^{n} \cdot 2} - 1$, but it didn't helped much. I am kind of stuck and don't know what to do next

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    $\begingroup$ Are you familiar with the difference of squares identity? $a^2-b^2=(a+b)(a-b)$... $\endgroup$ – abiessu Feb 10 at 21:11
  • $\begingroup$ @abiessu yes, but I don't see how it would help here. $\endgroup$ – Victor Feb 10 at 21:14
  • $\begingroup$ So, in this case, take $a=2^{2^n}$ and $b=1$, this gives that $2^{2^{n+1}}-1=(2^{2^n}+1)(2^{2^n}-1)$. What prime can divide both portions of this multiplication? $\endgroup$ – abiessu Feb 10 at 21:18
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    $\begingroup$ @abiessu Thank you so much, I got it. $\endgroup$ – Victor Feb 10 at 21:20
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    $\begingroup$ @Charlie FYI, the proof to your question is given as "Proof 2" in the question text of Different ways to prove there are infinitely many primes?. $\endgroup$ – John Omielan Feb 10 at 21:23
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As guided in the comments, taking $2^{2^n}-1$ as a difference of squares allows for the $n+1$ term to be written as

$$2^{2^{n+1}}-1=(2^{2^n}+1)(2^{2^n}-1)$$

On the RHS, any prime which divides both of the factors must divide their difference, which is $2$. But both factors are odd, so these factors are relatively prime to one another.

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