If 2^n + 1 is prime then n is a power of 2

Statement

Let $n$ be a positive integer. If $2^n + 1$ is a prime number, then $n = 2^k$ for a natural number $k$. So every prime of the form $2^n + 1$ with $n \ge 1$ is a Fermat number.

Remarks

The proof: if $n = k m$ with $m > 1$ odd, then $x + 1$ divides $x^m + 1$ for $x = 2^k$. So $2^k + 1$ is a proper divisor of $2^n + 1$. For example, $2^6 + 1 = 65 = 5 \cdot 13$, and $5 = 2^2 + 1$. The converse is false, because $F_5 = 2^{32} + 1$ is composite.

Explanations

L1 human

An odd factor of the exponent gives a factorization of $2^n + 1$. So the only exponents that can give primes are the powers of $2$, which have no odd factor except $1$.