Q:

Let $p$ be prime number, and d is the natural number. Prove that if $d\mid 2^p−1$, then $p\mid d−1$

Let $p$ be prime number, and d is the natural number. Prove that if $d\mid 2^p−1$, then $p\mid d−1$.
I'm looking on proof number 3 mentioned there and  few things are unclear for me: https://en.wikipedia.org/wiki/Mersenne_prime
How it was concluded that $q$ is a factor of $2^{pc}-1$ for all positive integers $c$. Fermat theorem says that $p\mid a^{q-1}-1$ if $q$ is not a divisor of $a$, in this case, let $b=a^c$ and it's true indeed, but in case of this proof exponent is $p$, not $p-1$. 
Second thing is why is there mentioned, that $q$ is not a factor of $2^1-1$ and how it leads to conclusion that $p$ is smallest positive integer $x$ that $q$ is a factor of $2^x-1$.

A:

We know that $2^p\equiv1\pmod q$. But then, for any natural number $c$,$$2^{pc}=(2^p)^2\equiv1^c=1\pmod q.$$
Now, let $n$ be the smallest natural number such that $q\mid 2^n-1$. You know that $n\neq1$. Suppose that $n\nmid p$. Then $p$ can be written as $nc+r$, for some $r\in\{1,2,\ldots,n-1\}$. But then$$1\equiv2^p=2^{nc+r}\equiv2^r\pmod q,$$which is impossible, since $n$ is the smallest natural with that property and $r<n$. So, $n\mid p$. Since $p$ is prime and $n\neq1$, $n=p$.

