Prove operator has eigenvalue with fundamental theorem of algebra

You know that $\sum_k a_k T^k v=0$ for some $a_k$. Define the polynomial $p(z)\equiv\sum_k a_k z^k$. You know that there are $\lambda_k$ and $c$ such that $p(z)=c\prod_k (z-\lambda_k).$ This also means that $p(T)=c\prod_k (T-\lambda_k I)$, by definition of $p(T)$.

I think what's important to note here is that you don't need $z$ to be a "number" for this decomposition to hold. If $z$ is, say, in $\mathbb C$, then you can think of the $\lambda_k\in\mathbb C$ as roots of the polynomial. If $z=T$ is an operator, then the $\lambda_k$ are not the roots of the polynomial anymore, in the sense that you don't have $p(T)=0$. However, you can still decompose the polynomial in the same way.

Our starting point, $\sum_k a_k T^k v=0$, now implies that $p(T)v\equiv\sum_k a_k T^k v=0$, that is, $c\prod_k (T-\lambda_k)v=0$.

The only way for this to happen is that $(T-\lambda_j)$ has a non-empty kernel for some $j$, which is equivalent to it being not injective.