Proof that the image of the adjoint representation is in the Lie algebra.

Question

I was reading a book on Symplectic Geometry and Analytical mechanics by Paulette Libermann. In the appendix she defines many basic things like Lie groups and Lie algebras and some things on them like the adjoint representation.
It's then said that for $g\in G,\,X\in\mathcal{G}$ the image of the adjoint representation $\mathrm{Ad}_gX:=(\alpha_g)_*(X)$ is in $\mathcal{G}$. Here $G$ is a Lie group, $\mathcal{G}$ its Lie algebra and $\alpha_g$ conjugation by $g$. It's also important to note that I consider an element of the Lie algebra to be a left invariant vector field here. So I tried to prove this by showing that $\mathrm{Ad}_gX$ is also a left invariant vector field. My proof is the following.\ Let $X\in\mathcal{G},\,f\in\mathcal{F}(G),\,g,g',x\in G$. Then\ \begin{align*} (L_{g'})_*(\mathrm{Ad}_gX(f))_x&=(L_{g'})_*((\alpha_g)_*X(f))_x\\ &=(L_{g'})_*((\alpha_{g^{-1}})^*\underbrace{X(f)}_{\in\mathcal{F}(G))_x})\\ &=(L_{g'})_*(X(f))_{\alpha_{g^{-1}}(x)}(\mathrm{d}_x\alpha_{g^{-1}})\\ &=((\underbrace{(L_{g'})_*X}_{=X})(f))_{\alpha_{g^{-1}}(x)}(\mathrm{d}_x\alpha_{g^{-1}})\\ &=(X(f))_{\alpha_{g^{-1}}(x)}(\mathrm{d}_x\alpha_{g^{-1}})\\ &=((\alpha_g)_*X(f))_x\\ &=(\mathrm{Ad}_gX(f))_x \end{align*} I am not quite sure about this however. Especially the step from the third row to the fourth row seems sketchy. Any critique is welcome and if anyone has a more concise or nicer way of showing this, I'd like to see it.