Factorial as Product of Consecutive Factorials

Theorem
The only factorials which are the product of consecutive factorials are:

Proof
Suppose $m, n \in \N$ and $m > n$.

Write $\map F {n, m} = n! \paren {n + 1}! \cdots m!$.

Suppose we have $\map F {n, m} > r!$ for some $r \in \N$.

Suppose further that there is a prime $p$ where $m < p \le r$.

We claim that $\map F {n, m}$ cannot be a factorial of any number.

$\map F {n, m} = s!$ for some $s \in \N$.

Since $s! > r!$, we must have $r! \divides s!$.

Since $p \le r$, we must have $p \divides r!$.

Thus we have $p \divides s!$.

However, since $n, n + 1, \dots, m < p$, we must have $p \nmid k!$ for each $n \le k \le m$.

Thus $p \nmid \map F {n, m} = s!$, which is a contradiction.

Hence $\map F {n, m}$ cannot be a factorial of some number.

We have the following lemmata:

Lemma 2
We also have:


 * $\forall n \in \N: n > 1 \implies \exists p: 2 n < p < 3 n$

Case $1$: $m$ is an even number larger than $2$
Write $m = 2 k$, where $k > 1$.

Then:

There is a prime $p$ where $m = 2 k < p \le 3 k - 1$.

Therefore $\map F {n, m}$ is not a factorial of some number.

Case $2$: $m$ is an odd number larger than $11$
Write $m = 2 k - 1$, where $k \ge 7$.

Then:

There is a prime $p$ where $m = 2 k - 1 < 2 k < p \le 3 k - 1$.

Therefore $\map F {n, m}$ is not a factorial of some number.

Case $3$: Particular values of $m$
The cases above leaves us with $m = 1, 2, 3, 5, 7, 9, 11$ to check.

We have:

Thus there are no more.