How can we possibly show this? We have such little control over anything. Wait, Does showing that \( 2^{ L f( 2^n)} \equiv 2^{ n - 2^n} \pmod { f(2^n)} \) remind us of anything? The Lefthandside makes it so tempting for us to want to apply Fermat's Little Theorem. Oh, dang! If only \(f(2^n)\) was a prime ...

Such wishful thinking. How do we "make" it a prime? Well, if it isn't a prime, how about we take a prime factor \(p\). Now, we backtrack our breadcrumbs to fix it. Remember when I said that breadcrumb 2 is too strong?

Breadcrumb 2B: For all \(n\), for any \( p \mid f(2^n)\), then there exists a \(k\) such that \( p \mid f(k)\) and \( p \mid f(2^k)\).

Breadcrumb 3B: Let's classify (possible) candidates for \(k\). From the 1st well known result, \(k = 2^n + Lp\), where \(L\) is any integer, work.

Breadcrumb 4B: We want to show that there is some \(L\) such that \( p \mid f(2^{ 2^n + L p})\).

**Exercise 8:** What do you think Breadcrumb 5B looks like?

Ponder this, and then move on to the next note in this set.

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