Why there are exactly $n$ distinct $n$-th roots?

Why there are exactly $n$ distinct $n$-th roots?

I was solving the following exercise: Given a complex number $z \neq 0$,
write $z = re^{i\theta}$ where $\theta = \arg(z)$. Let $z_1 =
Re^{i\alpha}$, where $R = r^{1/n}$ and $\alpha = \theta/n$, and let
$\epsilon = e^{2\pi i/n}$, where $n$ is a positive integer.
$(a)$ Show that $z_1^n = z$, that is $z_1$ is an $n$-th root of $z$.
$(b)$ Show that $z$ has exactly $n$ distinct $n$-th roots:
$$z_1, \epsilon z_1,\epsilon^2 z_1,\cdots, \epsilon^{n-1}z_1$$
Well, item $(a)$ was very easy indeed. Now letter $(b)$ is confusing
me.Showing that each of these is a $n$-th root was very simple. Indeed, if
we consider $\epsilon^k z_1$, then we have:
$$(\epsilon^k z_1)^n = \epsilon ^{nk}z_1^n = \epsilon^{nk}z=e^{i2\pi
k}re^{i\theta}=re^{i(\theta+2\pi k)} = re^{i\theta}$$
Where the last equality is because $k$ is integer, so $\cos(\theta + 2k
\pi) = \cos \theta$ and $\sin (\theta+ 2k\pi) = \theta$. I can't see why
there are exactly $n$ distinct roots, it seems to me that the problem is
that for $k > n-1$ the roots start to repeat, but I don't know how to show
it. For instance, for $k = n$, I cannot see why $\epsilon ^n z_1$ is a
repeated root. Is that the problem? Is really about the roots start
repeating? Can someone give a hint on how to show this?
Thanks very much in advance.