Semisimple Lie algebras are perfect.

Can anyone explain why a semi-simple finite dimensional Lie algebra $\mathfrak{g}$ has to be perfect ?

The natural way to prove something like that would be to look to the algebra generated by the Lie brackets, which when $\mathfrak{g}$ is not perfect would be expected to be solvable. But it doesn't seem to work.

Let $\mathfrak{g}$ be a semi simple Lie algebra. Then $\mathfrak{g}$ can be written as direct sum of simple ideals.

For simplicity, we can write $\mathfrak{g}=\mathfrak{g}_1\oplus\mathfrak{g}_2$ where $\mathfrak{g}_1,\mathfrak{g}_2$ are simple ideals.

We then have $$[\mathfrak{g},\mathfrak{g}]=[\mathfrak{g}_1\oplus \mathfrak{g}_2, \mathfrak{g}_1\oplus \mathfrak{g}_2]= [\mathfrak{g}_1,\mathfrak{g}_1]\oplus [\mathfrak{g}_2,\mathfrak{g}_2]=\mathfrak{g}_1\oplus \mathfrak{g}_2=\mathfrak{g}$$

See that $\mathfrak{g}_1=[\mathfrak{g}_1, \mathfrak{g}_1], \mathfrak{g}_2=[\mathfrak{g}_2,\mathfrak{g}_2]$ as $\mathfrak{g}_1,\mathfrak{g}_2$ are simple ideals. Thus, we have $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$.

One possible way (and I have not been able to think of an easier way) is the following:

Note that here I will assume this to be over $\mathbb{C}$.

That semisimple Lie algebras are perfect follows as a corollary to the theorem that such a Lie algebra is a direct sum of simple ideals. It is then clear as no non-trivial quotient of such a direct sum can be abelian.