Primes of form Power of Two plus One

Theorem
Let $$n \in \N$$ be a natural number.

Let $$2^n + 1$$ be prime.

Then $$n = 2^k$$ for some natural number $$k$$.

Proof
Suppose $$n$$ has an odd divisor apart from $$1$$.

Then $$n$$ can be expressed as $$n = \left({2 r + 1}\right) s$$.

But then:
 * $$2^{\left({2 r + 1}\right) s} = \left({2^s + 1}\right) \left({2^{2rs} - 2^{\left({2r-1}\right) s} + 2^{\left({2r-2}\right) s} - \cdots - 2^s + 1}\right)$$

as is apparent after some algebra. (Note that the result Sum of Odd Positive Powers‎ may also be invoked.)

Hence $$2^n + 1$$ can be prime only if $$n$$ has only even divisors.

That is, if $$n = 2^k$$ for some natural number $$k$$.

Note
Primes of the form $$2^{\left({2^k}\right)} + 1$$ are called Fermat primes.