6. Given that $n$ is an integer greater than 1, and $2^{2^{n}}+1$ is a prime number. Then $2^{2^{n}}-1$ must be ( ).
(A) a prime number
(B) a composite number with a factor of 3
(C) a composite number with a factor of 5
(D) a composite number with a factor of 15
6. (D).
$\because n$ is an integer greater than 1,
$$
\therefore N=2^{2^{n}}-1=16^{2^{n-2}}-1 \text {. }
$$

Therefore, the unit digit of $N$ is 5, so $5 \mid N$; $\because 2^{2^{n}}-1, 2^{2^{n}}, 2^{2^{n}}+1$ are three consecutive natural numbers, $\therefore$ one of them must be divisible by 3.
$\because 3 \times 2^{2^{n}}$, and $2^{2^{n}}+1$ is a prime number, $\therefore 3 \mid 2^{2^{n}}-1$. Hence, $2^{2^{n}}-1$ is a composite number with a factor of 15.