How to prove $\prod_{k=0}^n\left(2-\frac{2k+1}{n}\right)=-\frac{(2n)!}{2^n n^{n+1}n!}$.

Question

To finish a proof, I am stuck on the steps of getting from $$\prod_{k=0}^n \left(2-\frac{2k+1}{n}\right)$$ to the form $$-\frac{(2n)!}{2^n n^{n+1} n!}.$$ If it helps, the entire question as follows: we let $n\geq1$ and $x_j$ evenly spaced such that $x_j=(2j-n)/n$ for $j=1,\dots,n.$ Define $\phi_{n+1}(x)=(x-x_0)(x-x_1)\dots(x-x_n)$. With this, we want to show that $$\phi_{n+1}(1-1/n)=-\frac{(2n)!}{2^n n^{n+1} n!}.$$

I've written out and manipulated the values to arrive at what I think is the simplest form of the left hand side, as provided above, but I am not sure how to progress from where I'm at.

Answer 1

\begin{align} \prod_{k=0}^n \left(2-\frac{2k+1}{n}\right) &= \prod_{k=0}^n \left(\frac{2(n-k)-1}{n}\right) \\ &= \frac{1}{n^{n+1}}\prod_{k=0}^n (2(n-k)-1) \\ &= \frac{1}{n^{n+1}}\prod_{k=0}^n (2k-1) \\ &= -\frac{1}{n^{n+1}}\prod_{k=1}^n (2k-1) \\ &= -\frac{1}{n^{n+1}}\prod_{k=1}^n (2k-1) \frac{\prod_{k=1}^n (2k)}{\prod_{k=1}^n (2k)} \\ &= -\frac{\prod_{k=1}^{2n} k}{n^{n+1} 2^n \prod_{k=1}^n k} \\ &= -\frac{(2n)!}{2^n n^{n+1}n!} \end{align}