This seems a bit too straightforward but here goes: 2^(2^n) + 1 = a^5 - b^5 where a > b > 0. a^5 - b^5 = (a-b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4). So either a - b = prime and a^4 + a^3b + a^2b^2 + ab^3 + b^4 = 1 or a-b = 1 and a^4 + a^3b + a^2b^2 + ab^3 + b^4 = prime. In the first case we cannot have because a^4 + a^3b + a^2b^2 + ab^3 + b^4 clearly > 1. In the second case you have: a-b=1 implies b = a-1. So the factorization is: 2^(2^n) + 1 = 5a^4 -10a^3 + 10a^2 - 5a + 1 or 2^(2^n) = 5a^4 - 10a^3 + 10a^2 - 5a. But 5|RHS but not the LHS, contradiction. In fact it shows 2^k+1 is never a^5 - b^5. This stronger result makes me think i made an error somewhere, any mistakes?

I agree this was a really straightforward problem... Pierre.

Actually, if $p=a^5-b^5$, then $a,b$ are not multiles of $p$, so we can work in $\mathbb Z_p$ as follows: $(ab^{-1})^5-1=0\Rightarrow 5|p-1$. In our case this is $5|2^{2^n}$, clearly absurd. [I needed the fact that $b$ is not a multiple of $p$ so that $b^{-1}$ exists in $\mathbb Z_p$]

Nice solution grobber.

Hmmm....strange...none of the contestants did that Pierre.