There is  an alternative proof for FTA using "Fredholm operators on Hilbert spaces":

Assume that $P(z)=z^n+a_{n-1}z^{n-1}+\ldots+a_1 z+a_0$ has no root in $\mathbb{C}$. Then for  every $\epsilon$ the polynomial $Q(z)=\epsilon^nP(z/\epsilon)=z^n+\epsilon a_{n-1} z^{n-1} +\ldots+\epsilon^{n-1}a_1z+\epsilon^n a_0$ has no root in $\mathbb{C}$, too.

The space of entire  Holomorphic functions $Hol(\mathbb{C)}$ is densely  embedded in $\ell^2$ via $f(z)=\sum_{n=0}^{\infty} a_nz^n \mapsto (a_0,a_1,\ldots)$. We  substitute "$z$" in $Q(z)$ by the shift operator on $\ell^2$. Then  it turns out  that  every polynomial $Q(z) $  defines a  bounded linear operator $Q$ on $\ell^2$ which restricts to "multiplicative operator" by polynomial $Q(z)$ on $Hol(\mathbb{C})$ so the operator $Q$ keeps $Hol(\mathbb{C})$ invariant. Moreover a  non vanishing polynomial $Q(z)$ determines  an operator on $\ell^2$ which  restricts to a **surjective** operator on $Hol(\mathbb{C})$. 

Note that the operator $Q\in B(\ell^2)$ described above is  a  perturbation of the $n$-shift operator so $Q$ is a  fredholm operator of index $-n$. On the other hand it is  a  perturbation of the $n$-shift operator which is  an  isometry. So $Q$ satisfies $|Q(v)|>k|v|$ for  all $v \in \ell^2$ and for  some constant $k$. Then $Q$ is  an injective operator so it can not be a surjective operator  on $\ell^2$ since its index( $-n$ )is nonzero. On the other hand,  since  the restriction of $Q$ to a dense subspace of $\ell^2$ is surjective and  $Q$ satisfies $|Q(v)>k|v|$, then $Q$ must be  a  surjective operator on $\ell^2$, a contradiction.

**Added:** This proof is based on the following  note in arxiv however this arxived paper was not written well.(Full of typos, grammar problem, mistakes in english writing).

   https://arxiv.org/abs/math/0509113v1