Let $p$ be a prime and $q$ a prime divisor of $2^{p} -1$. Use Fermat's Little Theorem to prove that $q\equiv 1 (\mod \space p)$

Let $p$ be a prime and $q$ a prime divisor of $2^{p} -1$. Use Fermat's
Little Theorem to prove that $q\equiv 1 (\mod \space p)$

pQuestion continued: Hint: Consider $ord_{q}(2)$. Similarly, prove that if
$r$ is a prime factor of $2^{2^{k}}+ 1 $ then $r\equiv1 (\mod \space
2^{k+1})$/p pI think I have the first part, however I didn't really make
use of the hint, so I would greatly appreciate if someone could give me
some direction if my proof is not valid. I am also a little stuck on the
2nd part of the question, so if someone could give me some direction, it
would be greatly appreciated!/p pHere is my answer to the first part:/p
pWe have that $2^{p}\equiv1 (\mod \space q)$ and since $q$ is prime and
$q\nmid 2$,/p pby FLT, $2^{q-1}\equiv1 (\mod \space q)$/p pThen by
euclidean algorithm, we have that $d = ap + b(q-1) $ for some integers
$a,b$/p p$\implies 2^{d}\equiv1(\mod \space q)$/p pIf we assume $p ,(q-1)$
are not multiples of each other, then $d = ap + b(q-1) = 1$ (since $q$ is
prime and $q\nmid 2$ and $p$ is prime) /p p$\implies 2\equiv 1 (\mod
\space q)$, which is a contradiction./p pSo $p\mid (q-1) \implies
q\equiv1(\mod \space p) $/p pIs this proof correct? I have not used the
hint, but how would I go about doing that?/p pI wanted to use the same
strategy with the 2nd part of the question, and I have that:/p
p$2^{2^{k}}\equiv-1 (\mod \space r)$, and similarly by FLT, since $r$ is
prime and $r\nmid 2$,/p p$2^{r-1}\equiv1(\mod \space r)$/p pBut I am not
sure how to proceed after that. The $-1$ throws me off a bit. Any
direction would be greatly appreciated!!/p pThanks in advance!/p