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The topic of prime gaps is well-studied.

For this problem, for any number $n$ , set consider the sequence $n!+2, n!+3, \ldots, n!+n$ . The first term in the sequence is divisible by $2$ , since $2\mid n!$ and $2 \mid 2$ . Similarly, the second term in the sequence is divisible by $3$ , and so on (with the $k$ th term divisible by $k+1$ ) until we reach the last term, which must be divisible by $n$ . Thus, none of the numbers in our sequence can be prime, so these numbers belong to a gap of at least $n-1$ between two primes. So there must two consecutive primes that are separated by at least distance $d$ , for any value of $d$ .