Geometric progressions

Let $a$ be the first term, $r$ be the common ratio and $a_n$ be the required term. Let $a_n$ be the $n^{\text {th}}$ term of this GP.
$a_n = a × r^{n - 1}\\ 9! × {10}^{10} = 10! × {10}^{n - 1}\\ 9! × {10}^{10} = 9! × 10 × {10}^{n - 1}\\ {10}^{10} = {10}^n\\ \therefore n = \color{#69047E}{\boxed {10}}$

---

Generalizing:-
Consider a GP which starts with $n!$ has a common ratio of $n$ . Let $(n - 1)! × n^n$ be the $m^{\text{th}}$ term of this GP.
$a_m = a × r^{m - 1}\\ (n - 1)! × n^n = n! × n^{m - 1}\\ (n - 1)! × n^n = (n - 1)! × n × n^{m - 1}\\ n^n = n^m\\ \implies m = n$ .

So, the $(n - 1)! × n^n$ is the $n^{\text{th}}$ term of a GP which starts with $n!$ having a common ratio of $n$ .

In G.P. $n^{\text{th}}$ is given by $ar^{n-1}$ , where $a$ is first term and $r$ ia common ratio.

Given: $a=10!$ and $r=10$ .

Let the $n^{\text{th}}$ be $9!×10^{10}$ .

$\Rightarrow ar^{n-1}=9!×10^{10}$

$10!×10^{n-1}=9!×10^{10}$

$10^{n-1}=10^{-1}×10^{10}$

$n-1=10-1$

$n=\boxed{10}$