Trying to understand why "long blocks between primes" must exist I'm currently going through "An Introduction to the Theory of Numbers" by Hardy and Wright and at one point, they discuss why the distance from one prime to the next must have a long chunk of composites in between. I'm trying to understand the reasoning but I'm having some difficulty. Here is what they say:

Suppose that $2, 3, \ldots, p$ are the primes upto $p$. Then all the numbers up to $p$ are divisible by one of these primes, and therefore, if $2\cdot3\cdot5 \cdots p = q$ all of the $p-1$ numbers i.e. $q + 2, q+3, q+4 \ldots , q+p$ are composite.

I'm having particular difficulty understanding the $q + 2, q+3, q+4, \ldots , q+p$ part. If $q$ is equal to the product of primes from $2$ to $p$ then how can how can we have all the $p-1$ numbers be equal to $q+2, q+3, q+4, \ldots q + p$?
Can someone please clarify this? or if not, just explain what they are trying to say here. 
Thanks a bunch!
 A: Let's translate the case $p=7$:

Suppose that $2,3,5,7$ are the primes up to $7$. Then all the numbers up to $7$ are divisible by one of these primes, and therefore, if $2\cdot3\cdot5\cdot7=210$ all of the six numbers $212, 213, 214, 215, 216, 217$ are composite.

Hopefully, the above made sense. The authors are talking about the numbers $212, 213, 214, 215, 216,$ and $217$. There are six numbers listed, because $7-1=6$.
Grammatically, the clause "$p-1$" is being used to modify the plural noun "numbers". The authors are not equating, or comparing, any other variables with $p-1$.
A: They are saying that 
$q+2$ is divisible by $2.$
$q+3$ is divisible by $3.$
$q + 4$ is divisible by $2.$
And so on.
A: Let $n > 0$ then $ n! + 2, .... n! + n$ are all non-prime.  You can create a nonprime streak of arbitrary length using this process.
A: Psychology is such that sometimes things are hard when they are too simple. Possibly, a slightly stronger theorem + an exercise can help understanding.
Let $\ A\ $ be an arbitrary natural number (positive integer.
Then prove that all consecutive integers:
$$ A\!\cdot\!q+2\quad A\!\cdot\!q+3\quad \ldots
                                \quad A\!\cdot\!q+p $$
are composites -- thus, there are $\,\ p-1\,\ $ different composites which are consecutive integers each time (for each $\ A$).
As another mini-exercise, show that if $\ p>3\ $ then
also
$$ A\!\cdot\!q-2\quad A\!\cdot\!q-3\quad \ldots
                                \quad A\!\cdot\!q-p $$
are all composites.

WARNING. Without assumption $\ p>3\ $ the above mini-exercise fails (but only a little bit :) ).

