Why do we subtract 1 when calculating permutations in a ring? 
$10$ persons are to be arranged in a ring shape. Number of ways to do that is $9!.$

I wonder why we subtarct $1$ in all such cases.
I can imagine that if A,B,C,D are sitting in a row then B,C,D,A would give me a different combination but had they been sitting on a circular table then both the above combinations would imply the same thing. Same for CDAB or DABC.
But how does it all lead to $(n-1)!$ formula? Individual cases, I am able to imagine but overall how are we generalizing it?

Here's another way to think about it:
First, let's consider the case of the 'regular' linear permutation and see how we come up with the formula ${P(n) = n! }$. 
We have $n$ ordered slots (first, second, third etc.) in which we will insert our $n$ elements. Arbitrarily select an element, $i_1$. We have $n$ available slots in which we can insert $i_1$. Now choose another element, $i_2$. Independent of our choice for $i_1$, we have $n - 1$ options for $i_2$. Proceeding in this manner, we end up with ${n * n-1 * \ldots * 2 * 1 = n!}$.
Now let's consider the circular permutation. The ordering of a circle does not have a notion of a first element, second element etc. Rather we can describe the ordering of a circle in terms of element adjacency. In other words, we've described the configuration of a circle, when we've described for each element which other element lies to its right.
So let's choose one element, $i_1$. We have $n - 1$ potential right neighbors to choose from. Call our choice $i_2$, there are $n - 2$ options for $i_2$. (The element $i_1$ is not an option as a right neighbor for $i_2$  because that would prevent us from adding additional elements to the circle without breaking existing choices.) Proceeding in this way, our choices are ${n -1 * n -2 * \ldots * 2 * 1 * 1}$ (The last elements only have one option.) This gives the formula ($CP$ = number of circular permutations) ${CP(n) = (n-1)!}$. 
