TITLE: the sequence $n!+2,...,n!+n$ is made up of only composite numbers
QUESTION [5 upvotes]: I have found the claim that given that $n\geq2$, we have that the sequence of $n-1$ numbers $n!+2,n!+3,...,n!$ is made up of only composite numbers. Is there a proof of this? I found this pretty fascinating but I am not sure how to go around it. It seems to hold for the first few examples
$n=2$:  $S=\{4\}$
$n=3$: $S=\{8,9\}$
$n=4$: $S=\{26,27,28\}$

REPLY [2 votes]: I feel like I've answered this question before.
So $n!$ is divisible by $n$, right? Then $n! + n$ must also be divisible by $n$. This much should be obvious.
Since $$n! = \prod_{i = 1}^n i,$$ $n!$ must also be divisible by $n - 1$. In fact, $$\frac{n!}{n - 1} = (n - 2)! n.$$ Then $$\frac{n! + n - 1}{n - 1} = \ldots$$ Sorry, I forgot where I was going with that, I'm rushing to answer before this question gets closed on account of being a duplicate.
The point is that $n! + n - 1$ must be divisible by $n - 1$. And so we keep going down until to we conclude that both $n!$ and $n! + 2$ are even numbers. $n! + 1$ is divisible by $1$, but primes are divisible by $1$, so taht doesn't tell us much