Does a left-invariant vector field on a complex Lie group preserve holomorphic functions?

Let $G$ be a (finite-dimensional) complex Lie group, and suppose $f : G \to \mathbb{C}$ is holomorphic. Let $X$ be a left-invariant vector field on $G$. Must $Xf$ be holomorphic?

I think I have a proof, but I feel that I may have missed some subtlety, or overcomplicated matters. I'm still learning this area and I don't think my intuition is completely calibrated yet. I would also be happy to see alternate proofs.

Let $\mathfrak{g}$ be the Lie algebra of $G$, considered as the tangent space of $G$ at the identity $e$. Then $\mathfrak{g}$ has a complex structure; call it $J$. Without loss of generality we can assume $X$ is a real vector field, so let $\xi = X(e) \in \mathfrak{g}$.

By left-invariance it is sufficient to show $Xf$ is holomorphic at $e$. Fix a holomorphic system of coordinates $(z^1, \dots, z^n)$ in a neighborhood of $e$. I claim $\displaystyle \frac{\partial}{\partial \bar{z}_j} f = 0$ at $e$. Let $\bar{Z}$ be a left-invariant complex vector field with $\bar{Z} = \displaystyle \frac{\partial}{\partial \bar{z}_j}$ at $e$. Then $\bar{Z}$ is of type (0,1) so we can write $\bar{Z}(e) = \eta + i J \eta$ for some $\eta \in \mathfrak{g}$.

Now we have $(\bar{Z}Xf)(e) = (X \bar{Z} f)(e) + ([\bar{Z}, X]f)(e)$. Since $f$ is holomorphic and $\bar{Z}$ is of type (0,1), $\bar{Z} f = 0$ so the first term vanishes. The second term equals $[\eta + iJ\eta, \xi]f = ([\eta, \xi] + i J [\eta, \xi]) f$ which also vanishes since $[\eta, \xi] + i J [\eta, \xi]$ is also of type (0,1).

As before, assume without loss of generality that $X$ is real, let $\xi = X(e) \in \mathfrak{g}$ and let $\exp : \mathfrak{g} \to G$ be the exponential map. Let $f_t(g) = f(g \exp(t \xi))$ and recall that $Xf(g) = \frac{d}{dt}\Big|_{t=0} f_t(g)$. Since the group operation is holomorphic, $f_t$ is a one-parameter family of holomorphic functions, hence its derivative at $t=0$ is also holomorphic. (For instance, we can interchange derivatives with respect to $t$ and $\bar{z}_j$ to see that $Xf$ satisfies the Cauchy-Riemann equations.)