prime of the form $n!-(n+1)^9$?

Question

Is there a prime of the form $$n!-(n+1)^9$$ where $\ n\ $ is a positive integer ?

Motivation : For $\ k=1,2,3,4,5,6,7,8\ $ , there exists a positive integer $\ n\ $ such that $\ n!-(n+1)^k\ $ is a prime number. The smallest $\ n\ $ is $\ 3,10,10,12,12,66,18,18\ $ respective.

For $\ k=9\ $ , $\ n=2\ $ and $\ n=4\ $ give negative prime numbers , but apart from that , there is no $n\le 5\ 000$ doing the job.

Answer 1

Comment: probably this idea helps:

Wilson's theorem says:

$(p-1)+1\equiv 0 \bmod p$

Case 1-If (n+1)=q is prime we can write:

$[(n+1)-1]!+1\equiv 0\bmod (n+1)$

We rewrite $n!-(n+1)^k$ as:

$P=n!-(n+1)^K\equiv -1\bmod (n+1)+(n+1)^k\equiv -1 \bmod (n+1)$$\space\space\space\space (1)$

Case 2- If (n-1) is prime we have:

$[(n-1)-1]!+1+1\equiv 1\bmod (n-1)$

$P=1\mod (n-1)-(n+1)^k\equiv [1-(n+1)^k]\bmod (n-1)=m(n-1)-(n+1)^k+1$

considering the fact that:

$n!>(n+1)^{\lfloor \frac{n-1}2 \rfloor}$; if n is odd.

$n!>(n+1)^{\lceil \frac{n-1}2 \rceil}$; if n is even.

Substituting n by (n-1) we get:

$(n-1)!>n^{\lceil \frac{n-2}2 \rceil}$; if n is odd.

$(n-1)!>n^{\lfloor \frac{n-2}2 \rfloor}$; if n is even.

Suppose n is even and $n+1$ must be prime as OP suggest , then problem reduces to:

$$\begin {cases}P=m(n+1)-1; m,intger \\(n-1)!>n^{\lfloor \frac{n-2}2 \rfloor}\end{cases}$$

Update: The application of this theorem to relation (1) may also help:

Lioville theorem:

For prime number $p>5$ and natural number $m$ the equality $(p-1)!+1=p^m$ is not possible.

I saw this theorem and it's proof in this book:

"250 problemes de theorie elementaire des number"

By W. Sierpinsky.