Q.Find the value of $n$ if

Factorial of a natural number $n$, written $n!$, is the product of the first $n$ natural numbers:

read "$n$ factorial". It satisfies the recursive relation $n!=n(n-1)!$ for $n>1$ (and further unwindings $n!=n(n-1)(n-2)!$, etc.), and by convention $0!=1$ — forced by substituting $n=0$ into $(n+1)!=(n+1)\,n!$, which gives $1!=1\times0!\Rightarrow0!=1$. This convention lets factorial be defined for every non-negative integer.

Small values: $1!=1,\,2!=2,\,3!=6,\,4!=24,\,5!=120,\ldots$ — factorials grow extremely fast ($22!$ already has 22 digits).

The double factorial $n!!$ multiplies only every other integer down to $1$ or $2$:

It is a genuinely different operation from applying factorial twice: $n!!\ne(n!)!$. …