Let $P_n$ denote the product of all the coefficients in the expansion of expansion of ${(1+x)}^n.$ If $\left(20\right)!P_{n+1}={21}^{20}{P}_n,$ then $n$ is equal to

Suppose 

$(1+x{)}^{n+1}=\sum _{k=0}^{n+1} B_k{x}^k,$ where $B_k{=}^{n+1}{C}_k$

and $(1+x{)}^n=\sum _{k=0}^n C_k{x}^k,$ where $C_k{=}^n{C}_k,$

We have 

$\frac{P_{n+1}}{P_n}={\overline{B}}_0\left(\frac{B_1}{A_0}\right)\left(\frac{B_2}{A_1}\right)\cdots \left(\frac{B_{n+1}}{A_n}\right)$

Now, $\frac{B_{r+1}}{A_r}=\frac{(n+1)!}{(r+1)!(n-r)!}\cdot \frac{r!(n-r)!}{n!}$

$=\frac{n+1}{r+1} (0\le r\le n)$

Thus, $\frac{P_{n+1}}{P_n}=\frac{(n+1{)}^n}{r!}$

$\Rightarrow \frac{{21}^{20}}{20!}=\frac{(n+1{)}^n}{r!}\Rightarrow n=20$