An exam paper has given this question.
Let $k$ be a positive integer. We have to prove that $$(k+1)!+2, (k+1)!+3,...,(k+1)!+k, (k+1)!+(k+1)$$ are $k$ consecutive composite intergers.
All I need a proof verification.
Let $k>0$. Let us assume that, $$(k+1)!+2, (k+1)!+3,...,(k+1)!+k, (k+1)!+(k+1)$$ are $k$ consecutive prime numbers. Hence $k>1$.
Hence from $(k+1)!+2$ we get, $$(k+1)!+2 \\= \{(k+1).k\}.\{(k-1).(k-2)\}....4.3.2.1+2 \\ =2n.n'+2$$, [where $n=(k+1)k, n'=(k-1)!, M=n.n'+1$] $$2(n.n'+1)=2.M$$
Now, $M$ divides $(n.n'+1)$, $M$ divides $(k+1)!+2$, hence $(n.n'+1)$ divides $(k+1)!+2$. Hence, $(k+1)!+2$ is not a prime, it's composite.
Similarly, $(k+1)!+3$, $(k+1)!+4$ are composite. Let, $(k+1)!+k$ be composite but assume $(k+1)$ and $(k+1)!+(k+1)$ are prime.
By Wilson's theorem, $$(k+1-1)!\equiv -1\mod {k+1} \\ \Rightarrow k!\equiv k \mod {k+1} \\ \Rightarrow k(k+1)(k-1)!\equiv 0 \mod {k+1} \\ \Rightarrow (k+1)!+(k+1)\equiv (k+1)\equiv 0 \mod {k+1}$$
$(k+1)$ divides $(k+1)$ and $(k+1)!+(k+1)$ hence $(k+1)$ is composite, hence $(k+1)!+(k+1)$. Hence, $$(k+1)!+2, (k+1)!+3,...,(k+1)!+k, (k+1)!+(k+1)$$ are $k$ consecutive composite intergers.
Does it correctly complete the proof? Any suggestion or help is highly appreciated.
Your proof is too complicated.
The numbers are $(k+1)!+t$ with $2 \le t \le k+1$ and so $(k+1)!+t$ is clearly a multiple of $t \ge 2$, so not prime because the quotient is at least $2$:
$$ t \le k+1 \le (k+1)! \implies 2t \le (k+1)!+t \implies 2 \le \frac{(k+1)!+t}{t} $$