A complex polynomial of degree $n$ has exactly $n$ roots counted with multiplicity

By [L1] and [L2], there are $c\in {\mathbb{C}}^{\times }$ and complex numbers ${\beta }_1,\dots ,{\beta }_n$ such that $f(x)=c{\prod }_{k=1}^n(x-{\beta }_k)\mathrm{.}$ Let ${\alpha }_1,\dots ,{\alpha }_r$ be the distinct values among the ${\beta }_k$, and let $m_j$ be the number of indices $k$ with ${\beta }_k={\alpha }_j$. Then $f(x)=c{\prod }_{j=1}^r(x-{\alpha }_j{)}^{m_j},$ and by construction $m_1+\cdots +m_r=n$.

Suppose also that $f(x)=c^{\mathrm{\prime }}{\prod }_{j=1}^s(x-{\gamma }_j{)}^{n_j}$ with $c^{\mathrm{\prime }}\in {\mathbb{C}}^{\times }$, distinct ${\gamma }_j$, and positive integers $n_j$. In the UFD $\mathbb{C}[x]$, each linear factor $x-\alpha$ is irreducible, hence prime. Therefore the exponent with which $x-\alpha$ appears in a factorization of $f$ is uniquely determined. After reordering, this gives $r=s,{\gamma }_j={\alpha }_j,n_j=m_j\text{ for every }j\mathrm{.}$ So the multiplicities are well defined.

Step 1.1 gives a factorization whose exponents sum to $n$, and step 2.1 gives uniqueness of those exponents. This is exactly the statement that a degree-$n$ complex polynomial has exactly $n$ roots counted with multiplicity.