Prove Proposition (4.23).
You may find the result of Supplementary Problems 3, question 15 useful:
Let $\gcd (a, n) = 1$ and $k$ be the smallest positive integer such that $a^k \equiv 1 \pmod {n}$ .
Then $a^h \equiv 1 \pmod {n} \iff k \mid h$.
Let's remind ourselves of Proposition (4.23).
Let $q$ be an odd prime. Any prime factor $p$ of the composite Mersenne number $M_q = 2^q- 1$ is of the form $p= 2kq + 1$ where $k$ is an integer.
The smallest odd prime $q$ is 3, and so the smallest prime $p$ is 7, which means $\gcd(p,2)=1$. This allows us to use Fermat's Little Theorem (4.1) to give
$$ 2^{p-1} \equiv 1 \pmod p $$
Assuming $p$ is a factor of $2^q -1$ also gives us
$$ 2^q \equiv 1 \pmod p $$
We want to show $q$ is the smallest positive integer such that $2^q \equiv 1 \pmod p$. Let's assume there is another integer $s$ which is the smallest integer such that $2^s \equiv 1 \pmod p$. Using the given result, for some integer $j$, we have
$$ s \mid sj \quad \iff \quad 2^{sj} \equiv 1 \pmod p $$
Comparing this to $2^q \equiv 1 \pmod p$, noting that $q=sj$ is prime, means the smallest positive integer $s=q$. The other option $s=1$ is not possible because $2^1\equiv 1 \pmod p$ is a contradiction.
Having establishing $q$ is the smallest positive integer such that $2^q \equiv 1 \pmod p$ and comparing with $2^{p-1} \equiv 1 \pmod p$, the given result tells us
$$ 2^{p-1} \equiv 1 \pmod p \quad \iff \quad q \mid p-1 $$
That is, for some integer $n$,
$$ qn = p - 1 $$
Re-arranging
$$ p = qn + 1 $$
Since both $p$ and $q$ are odd primes, $n$ must be even, so we can write it as $n=2k$.
This gives us the desired conclusion
$$ p = 2kq + 1 $$