The statement (as in Bourbaki) is equivalent to: every (finite dimensional) Lie algebra with an invertible derivation is nilpotent.

Since every derivation of a semisimple Lie algebra is inner (this is elementary, see e.g. http://amathew.wordpress.com/2010/01/30/derivations-of-semisimple-lie-algebras-and-the-abstract-jordan-decomposition/), you already know the Lie algebra $\mathfrak{g}$ is solvable. Now from the derivation you can define a semidirect product $\mathfrak{h}=\mathfrak{g}\rtimes\mathfrak{a}$, where $\mathfrak{a}$ is the one-dimensional Lie algebra. Since the derivation is invertible, the derived subalgebra $\mathfrak{h}'$ is equal to $\mathfrak{g}$. Since the derived subalgebra of any solvable Lie algebra is nilpotent (consequence of Ado's Theorem [Edit: no it's not needed, see below]), it follows that $\mathfrak{g}$ is nilpotent.

[Edit: consider the adjoint representation of any solvable Lie algebra \mathfrak{h}; triangulate it (over an algebraic closure), so that $\mathfrak{h}'$ is mapped to algebra of upper nilpotent matrices. Since the adjoint representation has central kernel, this shows that $\mathfrak{h}'$ is nilpotent. Also as pointed out by Jim, I assume characteristic zero in the whole argument.]