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Additions in $color{blue}{textrm{blue}}$

$color{blue}{textrm{Suppose that}~ z^n-1 = 0~ textrm{has}~m > n~textrm{solutions}~ a_1, a_2, ldots, a_m}$.

Suppose that $a_1$ is a solution of $require{enclose}enclose{horizontalstrike}{z^n - 1 =0}$. Then:
$color{blue}{textrm{Consider the solution}~a_1~textrm{. Then:}}$

$$z^n-1 = (z-a_1)p_1(z) = zp_1(z)-a_1p_1(z),$$

where $p_1(z)$ is a polynomial. Since the degree of $zp_1(z)-a_1p_1(z)$ should be equal to $n$, then $p_1(z)$ has degree $n-1$.

Now, suppose that $require{enclose}enclose{horizontalstrike}{a_2}$ is another solution. Then:
$color{blue}{textrm{Now, consider the solution}~a_2.~textrm{Then:}}$

$$z^n-1 = (z-a_1)(z-a_2)p_2(z)=(z^2 ldots)p_2(z).$$

This time, $p_2(z)$ should have degree $n-2$.

In general, given $k$ solutions $a_1, a_2, ldots, a_k$, we ca write:

$$z^n-1 = p_k(z)prod_{i=1}^{k}(z-a_k),$$

where the degree of $p_k(z)$ is $n-k$. Of course, this can be reiterated up to $p_n(z)$, which has degree $n-n = 0$, i.e. it is $p_n(z) = p,$ a constant. Formally:

$$z^n - 1 = pprod_{i=1}^{n}(z-a_n).$$

Then, you have $n$ solutions $a_1, a_2, ldots, a_n$. Of course, if some $a_i$ coincide, then you have at most $n$ solutions.

$color{blue}{textrm{In conclusion, the number of solutions cannot be}~m>n, textrm{but}~ mleq n}$.

By the factor theorem , $f(z)$ is divisible by $(z-r)$ for each root $r$.

If there were more than $n$ roots, $r_1,dots,r_k$, then $f(z)=p_k(z)(z-r_1)dots(z-r_k)implies operatorname{deg}fge kgt n$.

Because it is a non-constant, single-variable, polynomial with complex coefficients of degree $n$ and the fundamental theorem of algebra says it has $n$ roots in $mathbb{C}$. Another way to say that is complex numbers is algebraically closed. By a successive factorization argument you can then show that the polynomial has exactly $n$ roots.

There's an explanation if you represent them in polar co-ordinates and consider that multiplying two complex numbers involves adding their arguments (ie angles) and multiplying their magnitudes (distances from the origin). It turns out they need to have $1$ as their magnitude and be multiples of$frac{360°}{n}$ apart on the resulting circle, so only $n$ of them will fit.

(This is effectively the same as gimusi's answer.)

Edit: In fact, there are exactly $n$ of them, and they're equally spaced round the circle. The reason should be obvious if you pick one of the candidate angles and multiply it by $n$.

First fact: for complex (nonzero) polynomials $f(z)$ and $g(z)$, the degree formula holds:
$$
deg(f(z)g(z))=deg f(z)+deg g(z)
$$
where $deg$ denotes the standard polynomial degree.

Proof. If we write
$$
f(z)=az^m+f_0(z),qquad g(z)=bz^n+g_0(z)
$$
where $f_0$ and $g_0$ group together the lower degree terms, $ane0$ and $bne0$, then
$$
f(z)g(z)=abz^{m+n}+h(z)
$$
where again $h(z)$ has degree less than $m+n$. Thus $f(z)g(z)$ has degree $m+n$. QED

Second fact (basic and well known: if $a$ is a root of the polynomial $f(z)$, then $f(z)$ is divisible by $z-a$.

Now we prove by induction on the degree of $f(z)$ the following statement.

Let $f(z)$ be a nonzero polynomial with coefficients in $mathbb{C}$. Then the number of distinct roots of $f$ cannot exceed the degree of $f$.

The statement is obvious for polynomials of degree $1$. Assume we know it for polynomials of degree $n-1$. Let $f(z)$ have degree $n$ and let $a_1,a_2,dots,a_m$ be its pairwise distinct roots. By the second fact, we have $f(z)=(z-a_m)g(z)$ and $g(z)$ has degree $n-1$. Now, for $k=1,dots,m-1$,
$$
f(a_k)=(a_k-a_m)g(a_k)=0
$$
and, since $a_k-a_mne0$, we conclude $g(a_k)=0$. Therefore $a_1,dots,a_{m-1}$ are pairwise distinct roots of $g(z)$. By the induction hypothesis, we have
$$
m-1le n-1
$$
and therefore $mle n$. QED

Note. This proof applies with no change to polynomials having coefficients in an arbitrary domain.