the sequence $n!+2,...,n!+n$ is made up of only composite numbers 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
$n=2$:  $S=\{4\}$
$n=3$: $S=\{8,9\}$
$n=4$: $S=\{26,27,28\}$

$n!$ is a rather special integer. It is composed of 1 factor of each from the list $\{1,2,3, \cdots, n-1, n\}.$ 
So,
$$n \geq 2 \Rightarrow 2|n! \Rightarrow n!+2 \equiv 0 \mod(2)$$
$$ n \geq 3 \Rightarrow 3|n! \Rightarrow n!+3 \equiv 0 \mod(3)$$
$$\vdots$$
$$ n \geq k \Rightarrow k|n! \Rightarrow n! + k \equiv 0 \mod(k)$$
$$\vdots$$
$$ n \geq n-1 \Rightarrow n-1|n! \Rightarrow n! + (n-1) \equiv 0 \mod(n-1)$$
$$n \geq n \Rightarrow n|n! \Rightarrow n! + n \equiv 0 \mod(n).$$
The last two steps shown here are rather obvious but I want to make it reasonably clear why every term in the list is composite.
Note that one of the nice things about this is that we have shown that there exists arbitrarily long lists of consecutive composite integers!
