Why does $z^n-1=0$ have at max n solutions? $z\in\mathbb{C}$ I know that there is a Theorem which says that a Polynom of Degree n has at most n Solutions, however we have not proved it yet in our class. Is there Maybe another explaination for this Special case?

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, $a\ne0$ and $b\ne0$, 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 $a_k-a_m\ne0$, 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
$$
m-1\le n-1
$$
and therefore $m\le n$. QED
Note. This proof applies with no change to polynomials having coefficients in an arbitrary domain.
