Proof $\prod_{i = 1}^n \frac{n + i}{2i-3} = 2^n(1-2n)$ using inducction

Question

i'm trying to solve this, using induction. The base step is easy, there's no difficult there.

The problem comes in the inductive step, I got to demonstrate that:

$$\prod_{i = 1}^{n+1} \frac{n+ 1 + i}{2i-3} = 2^{n+1}(1-2n -2)$$

Now, starting from the left side of the equation I quickly run into a problem. The extra $+1$ in $\frac{n+ 1 + i}{2i-3}$ won't let me use the inductive hypothesis. How can i solve this?

Answer 1

Hint:

$$\prod_{i=1}^{n+1} \frac{n+i+1}{2i-3}=\frac{\prod_{i=1}^{n+1}(n+i+1)}{\prod_{i=1}^{n+1} (2i-3)}$$

Now shift the index in the numerator: $\prod_{i=1}^{n+1} (n+i+1)=\prod_{i=2}^{n+2} (n+i)=?$