If $f$ is an analytic function in $z=0$ and $g(z)=f(z^2)$, prove that $g^{(2n-1)}(0)=0$ for all $n\in\mathbb{Z}^+$.

Question

I am trying to solve this problem:

If $f$ is an analytic function in $z=0$ and $g(z)=f(z^2)$, prove that $g^{(2n-1)}(0)=0$ for all $n\in\mathbb{Z}^+$.

My attempt: Since $f$ is analytic in $z=0$, $f$ has the power series representation $$f(z) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}z^n.$$ Therefore, $$g(z) = f(z^2) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}z^{2n}.$$ Because the Taylor series of $g$ around $0$ only contains even powers of $z$, we conclude that the odd derivatives of $g$ are zero.

Is this solution correct?