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I'm going through the proof that there exist numbers $n$ such that $n\,|\,2^n+1$ and $n$ has arbitrarly many distinct prime factors from the last post here.

The author defines $f(n)=2^n+1$, and, utilizing Lifting the Exponent lemma, proves that

$$\prod_{p|f(n)}p^{v_p(f(f(n)))}|f(n)^2.$$ Then he claims that as $m^2<2^m+1$ for $m\ge 4$, for $f(n)\ge 4$ it must be that $f(f(n))$ has a prime factor that doesn't divide $f(n)$.

I don't really seem why is that. Could somebody provide a clarification?

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If every prime appearing in $f(f(n))$ apears in $f(n)$, then $$f(f(n))=\prod_{p|f(f(n))}p^{v_p(f(f(n)))}|\prod_{p|f(n)}p^{v_p(f(f(n)))}|f(n)^2,$$ so, in particular, $f(f(n))\leqslant f(n)^2$. Now use your inequality for $m=f(n)$ to get a contradiction.

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