Calculate the distance between the points $(1, 2, \dots, n)$ and $(2, 3, \dots, n, 1)$ I know that the operation to find the distance between two vectors is:
$$\sqrt{(b_1-a_1)^2+(b_2-a_2)^2+...+(b_n-a_n)^2}.$$
So the distance between $(7, 5, 3, 1)$ and $(1, 3, 5, 7)$ is:
$$\sqrt{(1-7)^2+(3-5)^2+(5-3)^2+(7-1)^2} = \sqrt{80}.$$
The distance between $(1, 2, \dots, n)$ and $(0, 0, \dots, 0, 0)$ is:
$$\sqrt{(0-1)^2+(0-2)^2+\dots+(0-n-1)^2+(0-n)^2} = \|(1, 2,\dots, n)\|.$$
But what about the distance between $(1, 2, \dots,n)$ and $(2,3, \dots,n, 1)$?
$$\sqrt{(2-1)^2+(3-2)^2+\dots+(n-n-1)^2+(1-n)^2} = \sqrt{1 + 1 +\dots+ 1 + (1-n)^2} = ?$$
I think the result is $n-1$, because it adds one to one up to $\sqrt{(1-n)^2} = n-1$, but I'm not sure.
Someone can help me solve this?

Note that
$$\underbrace{1 + 1 + \dots + 1}_{n-1} + (1 - n)^2 = (n - 1) + (n - 1)^2 = (n-1)[1 + (n - 1)] = (n-1)n$$
where the first equality used the fact that 
$$(1 - n)^2 = [(-1)(n-1)]^2 = (-1)^2(n-1)^2 = (n-1)^2.$$
Therefore your calculation shows that 
$$\|(2, 3, \dots, n, 1) - (1, 2, \dots, n-1, n)\| = \sqrt{n(n-1)}.$$
