Exact functors and derived functors

Given an exact (additive) functor $$F$$, i.e. an additive functor preserving exact sequences, it is not hard to show that all derived functors of $$F$$ vanish.

At the same time given a right exact functor (a similar argument holds for the left exact case) one can show that for every short exact sequence $$0\longrightarrow A\longrightarrow B\longrightarrow C\longrightarrow 0$$ there exists an induced long exact sequence of the form $$\cdots \longrightarrow L_1F(B)\longrightarrow L_1F(C)\longrightarrow F(A) \longrightarrow F(B)\longrightarrow F(C) \longrightarrow 0$$ and hence if the first derived functor $$L_1F$$ vanishes, the functor is exact.

This seems to imply that the vanishing of the first derived functor is a sufficient condition for the vanishing of all higher derived functors. Is this true? This feels like a very strong result/constraint, so I get the feeling that I am missing something.

The following illustration may make this feel less surprising. Let $$A$$ be any object and take a short exact sequence $$0\to B\to P \to A\to 0$$ where $$P$$ is projective. There is then an induced long exact sequence $$\dots\to L_{n+1}F(P)\to L_{n+1}F(A)\to L_nF(B)\to L_nF(P)\to\cdots$$ When $$n\geq 1$$, $$L_nF(P)$$ and $$L_{n+1}F(P)$$ are trivial since $$P$$ is projective, and so the map $$L_{n+1}F(A)\to L_nF(B)$$ is an isomorphism. So, for instance, the vanishing of $$L_1F(B)$$ is equivalent to the vanishing of $$L_2F(A)$$. Iterating this construction, we can similarly find an object $$C$$ such that the vanishing of $$L_1F(C)$$ is equivalent to the vanishing of $$L_3F(A)$$, and so on. So if we know $$L_1F$$ vanishes on all objects, that actually tells us $$L_nF$$ vanishes on $$A$$ for all $$n\geq 1$$.