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 $f(z)=\sum a_iz^n \mapsto (a_0,a_1,\ldots)$. 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 on $\ell^2$ which restricts to "multiplicative operator" by $Q$ on $Hol(\mathbb{C})$ so it keeps $Hol(\mathbb{C})$ invariant. Moreover a non vanishing polynomial $Q$ determines an operator on $\ell^2$ which 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, that is the $n$-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 $\ell^2$ 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.