I have the following product:
$$\prod_{k=1}^{n+1}[2(n-k) +3] = (2[n-1]+3)(2[n-2]+3)(2[n-3]+3)....(2[n-n-1]+3)$$ which I can see now is equal to
$$1(3)(5)....(2n-3)(2n-1)(2n+1) = \prod_{k=1}^{n+1}(2k-1) = \prod_{k=2}^{n+1}(2k-1)$$
setting $k\rightarrow k+1$ gives $$\prod_{k=1}^{n}(2(k+1)-1) = \prod_{k=1}^{n}(2k+1)= \frac{(2n+1)!}{2^nn!}$$
I am just wondering if there is a way to use just the product notation to get
$$\prod_{k=1}^{n+1}[2(n-k) +3] = \prod_{k=1}^{n}(2k+1)$$
rather than expanding out the product in its terms and see that its true, is there some quick logical process that can be done on the first notation (product) that will instantly give me the last one or must I expand this out and undergo the process above. The equality of these two products may seem obvious but what about more complex products?
You can rewrite the equation likes the following:
$$2(n-k)+3 = 2(n-k)+ 2 + 1 = 2(n-k+1)+1= 2k'+1$$
It would be enough to find the range of $k' = n - k + 1$. As we know $k$ is from $1$ to $n+1$, we find $k'$ is changed from $0$ to $n$.