Prove that each member of the set of integers $n! +2 , n! +3,..., n! +n$ is divisible by a prime which doesn't divide any other members I am unable to solve this particular problem of Apostol introduction to analytic number theory on page 128 and I am self studying so I have no help.

Prove that each member of the set of n-1 consecutive integers n! +2 , n! +3,..., n! +n is divisible by a prime which doesn't divides any other member of the set.

I am unable to understand how can I choose that primes which divides 1 but not others. 2,3,4,...,n may be chosen but they are not all primes.
Can you please tell how should I approach this question.
thanks!!
 A: Look at the numbers $(n!+k)/k$ for $k=2,\dots,n$ and show that they are pairwise coprime.
By the way this gives an alternative proof of Euclid's theorem.
A: I think that Esteban Crespi's hint is helpful.
Let $j$ be an integer satisfying $2\leqslant j\leqslant n$, and consider $n!+j=j\bigg(\dfrac{n!}{j}+1\bigg)$ where $\dfrac{n!}{j}+1$ is an integer larger than $1$.
Now, let us prove the following two claims :
Claim 1 : If $k$ is an integer satisfying $2\leqslant k\leqslant n$ and $k\not=j$, then $\gcd\bigg(\dfrac{n!}{j}+1,k\bigg)=1$.
Claim 2 : If $k$ is an integer satisfying $2\leqslant k\leqslant n$ and $k\not=j$, then $\gcd\bigg(\dfrac{n!}{j}+1,\dfrac{n!}{k}+1\bigg)=1$.
From these claims, one can say that every prime factor of $\dfrac{n!}{j}+1$ is coprime to any other members of the set.

Proof for Claim 1 :
Since $k$ is an integer satisfying $2\leqslant k\leqslant n$ and $k\not=j$, one has $k\mid \dfrac{n!}{j}$ from which $\gcd\bigg(\dfrac{n!}{j}+1,k\bigg)=1$ follows.$\quad\blacksquare$

Proof for Claim 2 :
Suppose that there are integers $d(\geqslant 2),a,b$ such that
$$\dfrac{n!}{j}+1=da\qquad \text{and}\qquad \dfrac{n!}{k}+1=db.$$
Since $d$ cannot be divided by any integer $i$ satisfying $2\leqslant i\leqslant n$, one has $$d\geqslant n+1\tag1$$
One has
$$d|a-b|=|da-db|=\bigg|\bigg(\dfrac{n!}{j}+1\bigg)-\bigg(\dfrac{n!}{k}+1\bigg)\bigg|=\frac{n!}{jk}|k-j|$$
Since $\gcd\bigg(d,\dfrac{n!}{jk}\bigg)=1$, one gets $$d\mid |k-j|\tag2$$
It follows from $(1)(2)$ that
$$|k-j|\geqslant d\geqslant n+1$$
which contradicts that $j,k$ are integers such that $2\leqslant k\leqslant n$ and $2\leqslant j\leqslant n$.$\quad\blacksquare$

Added : This is an explanation about how the two claims prove what was asked in question.
Consider when $j=2$. Claim 1 shows that $\dfrac{n!}{2}+1$ is coprime to each of $3,4,\cdots, n$. Claim 2 shows that $\dfrac{n!}{2}+1$ is coprime to each of $\dfrac{n!}{3}+1,\dfrac{n!}{4}+1,\cdots, \dfrac{n!}{n}+1$. It follows from these that $n!+2=2\bigg(\dfrac{n!}2 +1\bigg)$ is divisible by a prime which doesn't divide any other members $n!+3=3\bigg(\dfrac{n!}3+1\bigg),n!+4=4\bigg(\dfrac{n!}4 +1\bigg),\cdots, n!+n=n\bigg(\dfrac{n!}n +1\bigg)$.
A: Here is one easy case:
Let $u=n!+a$ and $v=n!+b$.
If $\gcd(a,b)=1$, then $\gcd(u,v)=1$ because any prime divisor of $a$ divides $u$ but not $v$.
