# Factorial as Product of Consecutive Factorials

## Theorem

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

 $\ds 0!$ $=$ $\ds 0! \times 1!$ $\ds 1!$ $=$ $\ds 0! \times 1!$ $\ds 2!$ $=$ $\ds 1! \times 2!$ $\ds$ $=$ $\ds 0! \times 1! \times 2!$ $\ds 10!$ $=$ $\ds 6! \times 7!$

## 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.

Aiming for a contradiction, suppose $\map F {n, m} = s!$ 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 1

Let $n \in \N$.

Then $\paren {2 n - 1}! \, \paren {2 n}! > \paren {3 n - 1}!$ for all $n > 1$.

$\Box$

### Lemma 2

Let $n \in \N$.

Then $\paren {2 n - 2}! \, \paren {2 n - 1}! > \paren {3 n - 1}!$ for all $n \ge 7$.

$\Box$

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:

 $\ds \map F {n, m}$ $\ge$ $\ds \paren {m - 1}! \, m!$ $\ds$ $=$ $\ds \paren {2 k - 1}! \, \paren {2 k}!$ $\ds$ $>$ $\ds \paren {3 k - 1}!$ Lemma 1

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:

 $\ds \map F {n, m}$ $\ge$ $\ds \paren {m - 1}! \, m!$ $\ds$ $=$ $\ds \paren {2 k - 2}! \, \paren {2 k - 1}!$ $\ds$ $>$ $\ds \paren {3 k - 1}!$ Lemma 2

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:

 $\ds 10! \times 11!$ $>$ $\ds 13!$ and $13$ is a prime $\ds 8! \times 9!$ $>$ $\ds 11!$ and $11$ is a prime $\ds 6! \times 7!$ $=$ $\ds 10!$ $\ds 5! \times 6! \times 7!$ $>$ $\ds 11!$ and $11$ is a prime $\ds 4! \times 5!$ $=$ $\ds 2880$ is not a factorial of some number $\ds 3! \times 4! \times 5!$ $>$ $\ds 7!$ and $7$ is a prime $\ds \map F {n, 3}$ $=$ $\ds 12$ is not a factorial of some number $\ds \map F {n, 2}$ $=$ $\ds 2!$ $\ds \map F {0, 1}$ $=$ $\ds 1!$ $\ds$ $=$ $\ds 0!$

Thus there are no more.

$\blacksquare$