Prove that $2^{3^n}+ 1$ is composite for all natural numbers $n$.
[Hint: $x^m + 1 = (x + 1) (x^{m−1}− x^{m−2} + x^{m−3} − ⋯ + x^2− x + 1)$ , provided $m$ is odd.]
We can take $m=3^n$ as odd because $m$ only has prime factors 3, and not 2, so is not even.
This means we can use the given factorisation as
$$ 2^{3^n} + 1 = (2+1)( 2^{3^n-1} - 2^{3^n-2} + 2^{3^n-3} - \ldots + 2^2 -2 + 1) $$
The RHS is a product of two integers, and so $2^{3^n}+ 1$ is composite for all natural numbers $n$.
Update: the textbook author's solution requires us to check that $2^{3^n}+ 1$ is greater than one. This is because the book defines a composite as greater than 1. We can see this is the case because $3^n \ge 1$ for natural numbers $n$ (including $n=0$), and so $2^{3^n} + 1 \ge 2^1+1= 3$.