Two Fredholm operators A and B have the same index iff there is an invertible operator C s.t. A-BC is compact

Question

First, I've shown that a Fredholm operator $T\in B(E)$, where $E$ is an infinite dimensional Banach space, is a compact perturbation of an invertible operator iff its index vanishes.
Then, I need to show that two Fredholm operators $A,B \in B(E)$ have the same index iff there is an invertible operator $C$ such that $A-BC$ is compact.
If $A-BC \in K(E)$ and $C$ is ivertible then it follows easily that $ind(A)=ind(B)$.
For the other direction, I know few properties of the index:
Addition formula, stability of index under compact perturbations and continuity of index (given $A\in F(E)$, let $A_1,A_2,...\in B(E)$ be a sequence that converges to $A$, then there is $n_0$ s.t. for $n\ge n_0$, $A_n$ is a Fredholm operator with $ind(A_n)=ind(A)$.
I would like to have a hint, and not a full answer. Thank you!

Answer 1

I think I have a solution to my answer:
First, assume $ind(A)=ind(B)=0$. Then there exist $K_1,K_2 \in K(E)$ and $I_1,I_2$ invertible operators in $B(E)$ such that $A=K_1+I_1$ and $B=K_2+I_2$.
Define $C=I_2^{-1}I_1$, then $C$ is invertible and
$A-BC=K_1+I_1-(K_2+I_2)I_2^{-1}I_1=K_1-K_2I_2^{-1}I_1 \in K(E)$
as desired.
Now, for the general case: Assume $ind(A)=ind(B)=n$, then we can apply the stabilization trick, i.e. to look at $A\oplus T_{z^n}: E\oplus l^2 \to E\oplus l^2$ that have vanishing index.