Just a Conjecture

The numbers of the form $F_n = 2^{2^n}+1$ are known as fermat primes.

And it's a well-known fact (conjecture) that only $F_0,\ F_1,\ F_2,\ F_3,\ F_4$ are the only primes.

Infact, it still remains a open problem to prove that -

Is $F_n$ composite for all $n>4$

Also, it has been verified that $F_k$ is a composite number for $5 \leq k \leq 32$

We can show that 641 divides $2^{2^5}+1$ without doing a "brute force" calculation. Note that $641=2^4+5^4=5\times2^7+1$ and consider the following congruences modulo 641:
$2^{2^5}+1 =1+2^4\times2^{28}\equiv 1-5^4\times2^{28}=1-(5\times2^7)^4\equiv1-(-1)^4=0$

I reasoned the question wouldn't have been asked if the answer was true, since that would have been trivial. Saved a lot of factorization!

Kishlaya Jaiswal talked about the Fermat primes, and I'm going to add how the definition of Fermat primes can be strenghtened without making Fermat primes more difficult to handle and it is to show you that the definition of Fermat primes is not arbitrary -- the power of $2$ in the exponent may at first make you think that the definition is very specific and not useful.

Instead of primes of the form $2^{2^n}+1$ Fermat primes can be called primes of the form $2^m+1, m\in\mathbb Z^+$ , since $2^m+1, m\in\mathbb Z^+$ can only ever be prime if $m$ has no odd prime divisor --

If $m=p$ ( $p$ odd prime), then $2^p+1=3(2^{p-1}-2^{p-2}+\cdots+1)$ , where $(2^{p-1}-2^{p-2}+\cdots+1)$ is larger than $1$ for all odd primes $p$ , since $2^{p-1}-2^{p-2}=2^{p-2}>1$ .

If $m=kp, k\in\mathbb Z_{>1}$ ( $p$ odd prime), then $2^{kp}+1=(2^p+1)(2^{kp-p}-2^{kp-p-1}+\cdots+1)$ , where $(2^{kp-p}-2^{kp-p-1}+\cdots+1)$ is larger than $1$ for all odd primes $p$ and $k\in\mathbb Z_{>1}$ , since $2^{kp-p}-2^{kp-p-1}=2^{kp-p-1}>1$ .

Much simpler answer: any number raised ^5 ends with the same digit as the original number.

2^2^5 must end in a 4. Thus this result +1 this will make an answer with 5 as a factor so it won't be prime.

$2^{2^5}+1=4294967297=641\times 6700417$ so it is not a prime number.

The number to be prime has to be in the form of 6k+1 or 6k-1 so that mean 2^2^5 has to be divisible by 6 which if u find is not.. so the number is not prime. :)