Eigenstate of unitary operator used for Order-Finding

Question

In the "Quantum Computation and Quantum Information 10th Anniversary textbook by Nielsen and Chuang", chapter 5.3.1 introduces the concept of solving the Order-Finding Problem.

(Eqn 5.36) states the following:

$$U|y\rangle = |xy\text{ mod } N\rangle \text{ whereby } x<N \text{ and } \gcd(x,N)=1$$

(Eqn 5.37) defines the eigenstates of the unitary operator $U$ as:

$$|u_s\rangle = \frac{1}{\sqrt{r}}\sum_{k=0}^{r-1}\exp\left(\frac{-2\pi isk}{r}\right)\left|x^k\text{ mod } N\right\rangle \text{ for integer } 0 \leq s \leq r-1$$

Then when $U$ acts on $|u_s\rangle$, (Eqn 5.38 to 5.39) gives:

$$U|u_s\rangle = \frac{1}{\sqrt{r}} \sum_{k=0}^{r-1}\exp\left(\frac{-2\pi isk}{r}\right) |x^{k+1}\text{ mod } N\rangle = \exp\left(\frac{2\pi is}{r}\right)|u_s\rangle$$

I tried to work out Eqn 5.38 to 5.39 by substituting the index $k$ as dummy index $v=k+1$ which I get:

$$U|u_s\rangle = \exp\left(\frac{2{\pi}is}{r}\right) \frac{1}{\sqrt{r}}\sum_{v=1}^{r}\exp\left(\frac{-2\pi isv}{r}\right)|x^v\text{ mod }N\rangle$$

Question

Why is $\frac{1}{\sqrt{r}}\sum_{v=1}^{r}\exp\left(\frac{-2\pi isv}{r}\right)|x^v\text{ mod }N\rangle = |u_s\rangle$ when the summation limits are different? $|x^0\text{ mod }N\rangle$ and $|x^r\text{ mod }N\rangle$ are not the same right?

Answer 1

$$x^0 \bmod N = 1 \implies x^r \bmod N = 1$$ as by definition, $r$ is the order of $x \bmod N$ i.e. $r$ is the least integer that satisfies $x^r=pN+1$ for some integer $p$.