Binomial Theorem?

The function $f(x)$ has the form $f(x)=c(x-1)(x-2)...(x-n+1)$ , now we should solve for $c$ .

Plug in $x=n$ we obtain $n=f(n)=c(n-1)(n-2)...(1)=c(n-1)!$ , so $c=\dfrac n{(n-1)!}$

Now plug in $x=2n$ and you can solve for $n$ directly:

$923780=f(2n)=c(2n-1)(2n-2)...(n+1)=\dfrac n{(n-1)!}*\dfrac{(2n-1)!}{n!}=n*\dbinom{2n-1}{n}$ , so $n=10$