I was reading the following exercise:
Prove that if $n \ge 2\space$ then among the numbers: $n! + 2, \space n! + 3,..., n! + n$ none are prime (where $n! = 1\cdot 2 \cdot 3 \cdot ... n$
My approach:
Let $k \in {2, 3, 4...n}$
Then for any number $n! + k$ we have that $k | n! + k$
This is because $k | n!$ since $k$ is a factor of $n!$ (belongs to one of the products) and $k | k$.
So we have the positive integer: $$\frac{n! + k}{k} = \frac{n!}{k} + \frac{k}{k} = \frac{n!}{k} + 1 = c + 1$$ where $c = \frac{n!}{k}$
Now we know that $c$ is a positive integer, we also know that $c + 1 > 1$ and that $c \ne n! + k \space$ hence there is another number that divides $n! + k$ besides $1$ and $n! + k$ and so the number $n!+ k $ can not be a prime.
Is this a correct approach?
There has been extensive discussion regarding the answer in the comments, but nevertheless I will add an answer for the future readers.
The sequence of numbers you have consider are as follows
$$2!+2 \qquad n=2$$ $$3!+2,\space 3!+3 \qquad n=3$$ $$4!+2,\space 4!+3 ,\space 4!+4 \qquad n=4$$ $$\vdots$$ $$n!+2,\space n!+3,\space \cdots n!+k,\space\cdots, n!+n\qquad n=n$$
Now, we know, $n!=1\cdot2\cdot3\cdots n$. Clearly, for $k<n$, we can write $n!$ as $n!=1\cdots k\cdots n$ where $n\in\{2, 3, ..., n\}$. Hence we can write $n!+k$, where k is defined as before, as $$k\bigg((1\cdots(k-1)\cdot (k+1)\cdots n)+1\bigg)$$
Clearly, $k(<n)$ divides $n!+k$. Thus, none of $n!+2,\space n!+3 ,\space n!+(n-1)$ are prime because they have $k$ as a factor.
And for $n+n!$, the result follows almost trivially!