Show that for every $n > 1$ there exist $n$ consecutive composite numbers

Question

So I am trying to prove that for every $n > 1$ there exist $n$ consecutive composite numbers but I do not know even how to start. This is a problem in analytic number theory. Please can you help me solve this question in analytic number theory. Thank you

Answer 1

Consider the numbers:

$$(n+1)! + 2, (n + 1)! + 3, ..., (n + 1)! + n + 1$$

Answer 2

Another approach, using multiples of $2$ and $3$:

If $n$ is odd then $3|3n$ and $2 | 3n + 1$.

If $n$ is even, then $2 | 3n + 2$ and $3|3n + 3$.