Factorization of polynomial by Liouville Let's fix a complex polynomial $P\colon \mathbb{C} \to \mathbb{C} \mid P(z)=a_nz^n+a_{n-1}z^{n-1}+\dots+a_0$ with $a_n \neq 0$ and $n \ge 1$.
By Liouville I have proved that there exists a $z_0 \in \mathbb{C} \mid P(z_0)=0$.
Now I would like to show, as a corollary, that there exist $z_1,\dots,z_n \in \mathbb{C} \mid \forall \,z \in \mathbb{C} \quad P(z)=a_n(z-z_1)\cdots(z-z_n)$.
I started saying that, as said above, there exists $z_1 \in \mathbb{C} \mid P(z_1)=0$.
So we have $P(z)=P(z)-P(z_1)=\sum_{k=0}^n a_kz^k-\sum_{k=0}^n a_kz_1^k=\sum_{k=0}^n a_k(z^k-z_1^k)=$
$=\sum_{k=0}^n a_k[(z-z_1)\sum_{h=0}^{k-1} z^{k-1-h}z_1^h]=(z-z_1)\sum_{k=0}^n a_k\sum_{h=0}^{k-1} z^{k-1-h}z_1^h=(z-z_1)Q(z)$,
where $Q(z)=\sum_{k=0}^n a_k\sum_{h=0}^{k-1} z^{k-1-h}z_1^h$ is a new polynomial of degree $n-1$.
We have also used the identity $a^n-b^n=(a-b)\sum_{k=0}^{n-1} a^{n-1-k}b^k$.
We can repeat the procedure to $Q$ and so on, thus obtaining the result.
Am I correct? There exists a more direct way? Thank you!

Yes, you are right. In fact, what you have done is proving the polynomial remainder theorem or little Bézout's theorem which states that let $P(z) \in \mathbb{C}[P]$ with $\deg P = n$ has $z_0$ a root of it then there exists a polynomial $Q(z), \deg Q= n-1$ with complex coefficients and the leading coefficient is $1$ that: $P(z) = a_n (z-z_0)\cdot Q(z)$. By Liouville, there exists $z_1$ that is a root of $Q(z)$ and apply little Bézout's theorem again. Doing similarly for $n$ times yields your result.  that there exists $z_1,\dots,z_n \in \mathbb{C} \mid \forall \,z \in \mathbb{C} \quad P(z)=a_n(z-z_1)\cdots(z-z_n)$.
