That's Too Many Terms To Write!

Relevant wiki: Arithmetic Progressions

Using the formula $a_n = a + \left(n - 1\right)d$ .
$100! = 100 + (n - 1)×100\\ 100! - 100 = 100(n - 1)\\ \dfrac{100\left(99! - 1\right)}{100} = n - 1\\ 99! - 1 = n - 1\\ n = \color{#69047E}{\boxed{99!}}$

Generalizing:-
Let $n!$ be the $x^{\text{th}}$ term in an AP which starts with $n$ and have common difference between its terms are $n$ .
$n! = n + (x - 1)×n\\ n! - n = (x - 1)×n\\ x - 1 = \dfrac{n\left((n - 1)! - 1\right)}{n}\\ x - 1 = (n - 1)! - 1\\ x = (n - 1)!$

So, $n!$ is the ${(n - 1)!}^{\text{th}}$ term of an arithmetic progression starting with $n$ having common difference between its terms as $n$ .

Relevant wiki: Arithmetic Progressions

The $n^{\text{th}}$ term of an AP is given by $a_n = a + (n-1)d$ , where $a$ and $d$ is the first term and the common difference of the AP respectively. In this case, $a=d=100 \implies a_n = 100 + 100(n-1) = 100n$ . For $a_n = 100! = 99! \times 100 \implies n = \boxed{99!}$ .

The $nth$ term of an Arithmetic Progression is

$a_{n} = a_{1} + (n - 1)d$

Substituting gives

$100! = 100 + (n - 1)(100) = 100 + 100n - 100$

$100! = 100n$

Dividing both sides by $100$ gives

$n = 99!$

$ِa_{n}=a_{1} + (n-1)d = 100 + 100n-100 = 100! = 100(99!)$

$\because 100 + 100n-100 = 100! = 100(99!) \therefore 1+n-1=99!$

$\therefore n=99!$

Relevant wiki: Arithmetic Progression.

In A.P. $n^{\text{th}}$ term is given by $a_{n}=a+(n-1)d$ .

$\Rightarrow a+(n-1)d=100!$
$100(1+(n-1))=100×99!$
$\therefore n=\boxed{99!}$