For a positive integer $\ n\ $ , define $$f(n)=|\sum_{j=1}^n (-1)^j\cdot j!|=n!-(n-1)!+(n-2)!-(n-3)!\pm \cdots$$
I want to prove the
Conjecture : Every prime factor $\ p\ $ of $\ f(n)\ $ must satisfy $\ p>n\ $
My try : Assume $\ p\le n\ $ and $\ p\mid f(n)$. For every integer $\ k\ $ with $\ p\le k\le n\ $ we have $$f(k)+f(k-1)=k!\equiv 0\mod p$$ hence we can conclude $$p\mid f(p-1)$$ because of $$p\mid f(k)\implies p\mid f(k-1)$$ for $\ p\le k\le n\ $
But brute force reveals $\ n\nmid f(n-1)\ $ for $\ 2\le n\le 10^5$. But how can I prove $\ n\nmid f(n-1)\ $ for every $\ n\ge 2\ $ ? (it would be enough to prove it in the case that $n$ is prime) ? This is where I got stuck.
Comment:
Let's consider the first four terms, we have:
$A_n=(n-2)![n(n-1)+1]-(n-3)![(n-1)(n-2)+1]$
Therefore $n ł A$. For $A_{n-1} $ we have:
$A_{n-1}=(n-3)![(n-1)(n-2)+1]-(n-4)![(n-2)(n-3)+1]$
Therefore if n is odd prime then $n\not| A_{n-1}$
Subsequent groups of four terms have similar factors as first four terms and we may conclude that $n\nmid f(n-1)$ if n is odd prime.