A follow up problem

let there be a polynomial $p(x) = x f(x)-1$ . notice that $p(x)$ has degree $2n+1$ and roots $1,2,...,2n+1$ . hence $p(x)= a(x-1)(x-2)...(x-[2n+1])\to f(x) =\dfrac{a(x-1)(x-2)...(x-[2n+1])+1}{x}$ for some constant $a$ . for $f(x)$ to be a polynomial, we need $x \big|a(x-1)(x-2)...(x-[2n+1])+1 \to x \big| a(-1)(-2)...(-[2n+1])+1 \\ \to a(-1)(-2)...(-[2n+1])+1=0 \to a= \dfrac{1}{(2n+1)!}$ hence $f(x) =\dfrac{\frac{1}{(2n+1)!}(x-1)(x-2)...(x-[2n+1])+1}{x} \to f(2n+3) = \dfrac{\frac{1}{(2n+1)!}(2n+2)!+1}{2n+3}=\boxed{1}$