Given a group $G$,the derived subgroup of $G$ denoted by $[G,G]$ is defined as $[G,G]=\langle [a,b]:a,b \in G\rangle$ ,where $[a,b]$ is called the commutator of $a$ and $b$.
One can define: $G^{(0)}=G$ and for $n \in \mathbb N^+:G^{(n)}=[G^{(n-1)},G^{(n-1)}]$
Then the series $$...G^{(3)}\trianglelefteq G^{(2)}\trianglelefteq G^{(1)}\trianglelefteq G^{(0)}=G$$
Is called the derived series.
I'm trying to show that $\forall n \in \mathbb N^+:G^{(n)}\trianglelefteq G^{(n-1)}$ using induction on $n$.
Base case: By the definition $G^{(1)}=[G,G]$,take $a \in G$ and $b \in [G,G]$ then $[a,b]=aba^{-1}b^{-1} \in [G,G]$,since $[G,G]$ is a subgroup it implies that $[a,b]b=aba^{-1} \in [G,G]$ which shows that $G^{(1)}\trianglelefteq G^{(0)}=G$.
Now assume $G^{(n)}\trianglelefteq G^{(n-1)}$ does hold and consider $G^{(n+1)}$.
This is where do I have a problem and cannot finish the proof.
Just take $H=G^{(n-1)}$ and use your proof in the case $n=1$ to show $[H,H]=[G^{(n-1)}, G^{(n-1)}]=G^{(n)}$ is a normal subgroup of $H$. I think you are a bit confused about the notation but after you think about it the answer you are looking for is that, simply, the commutator subgroup of a group is normal.