Is it correct: $1!+2!+\ldots+n!$ is not divisible by $n+1$ for $n\ge3$?

Question

Is it correct: $1!+2!+\ldots+n!$ is not divisible by $n+1$ for $n\ge3$ ?

To me it seems that at least it is true for odd $n$.

Edit: The main question, I was dealing with is $1!+2!+\ldots+n!$ is not divisible by any prime $p\le n.$ What I did:

$$1!+2!+\ldots+n!=1!+2!+\ldots+(p-1)!+p!+\ldots+n!$$ So, if I can prove that $1!+2!+\ldots+(p-1)!$ is not divisible by $p,$ it is done. Taking some random $n,$ I didn't find any counter (my bad choices of $n$) and thought of the above conjecture which is not true as pointed in the below answers.

Answer 1

When $m>1$, then $m!$ is even. Therefore, $$ 1!+2!+3!+\dots+n! $$ is odd. When $n$ is odd, then $n+1$ is even. Therefore, $$ (n+1)\nmid(1!+2!+3!+\dots+n!) $$ since $2$ divides the LHS, but not the RHS.

On the other hand, when $n=8$, $$ 1!+2!+3!+4!+5!+6!+7!+8!=46233, $$ which is divisible by $n+1=9$.

Therefore, the claim is true when $n$ is odd, but it is false when $n$ is even. You may have been detecting that many of the first few odd numbers are prime.

Answer 2

Note that $1!+2!+\cdots+8!=46\,233=9\times5\,137$.