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 $Q(z)=\epsilon^nP(z/\epsilon)$ has no root in $\mathbb{C}$, too.

The space of entire  Holomorphic functions $Hol(\mathbb{C)}$ is densely  embedded in $\ell^2$ via $f(z)=\sum a_iz^n \mapsto (a_0,a_1,\ldots)$. Counting "$z$" as the shift operator on $\ell^2$, we realize that  every polynomial $Q(z) $  define a  bounded linear operator on $\ell^2$ which keeps $Hol(\mathbb{C})$ invariant. Moreover a  non vanishing polynomial $Q$ restricts to a **surjective** operator on $Hol(\mathbb{C})$. 

Note that the polynomial $Q$ described above is  a  perturbation of the $n$-shift operator so it is a  fredholm operator of index $-n$. On the other hand it is  a  perturbation of an isometry $n$-shift operator so it satisfies $|Q(v)>k|v|$ for  all $v \in \ell^2$ for  some constant $k$. Then $Q$ is  an injective operator so it can not be a surjective operator  on $\ell^2$. 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.