Evaluating the $k$th derivative of composite $g\circ f$ if $f^{\prime}(0) = 0$ without using Bruno's theorem

Question

Suppose $f$ and $g$ are infinitely differentiable. Suppose $f^{\prime}(0) = 0$. Consider the composite function $g \circ f$. If I wanted to calculate the $k$th derivative of $g \circ f$ and evaluate this function at $0$, we know a priori that it will be zero. However, to prove this without appealing to Bruno's Formula, would it be sufficient to observe that $$\frac{d^k}{dx^k} g(f(x)) = \frac{d^{k-1}}{dx^{k-1}} \big[g^{\prime}(f(x)) f^{\prime}(x)\big]$$ so that $$\frac{d^k}{dx^k} g(f(0)) = \frac{d^{k-1}}{dx^{k-1}} \,\big[g^{\prime}(f(0)) f^{\prime}(0)\big] = 0$$ Edit: Assuming that $0$ is in the domain of $f$, of course.

Answer 1

Counterexample: $g(x)=e^x$ and $f(x)=x^2$. If $h:=g \circ f$, then $h''(0)=2 \ne 0.$