Let $G$ be a group generated by: $\{g_1,...,g_n\}$, then $[G,G] \leq \langle g_1,...,g_{n-1}\rangle$ if and only if $\langle g_1,...,g_{n-1}\rangle$ is normal in $G$.
I think I have to prove that $gg_1g^{-1}\in \langle g_1,...,g_{n-1}\rangle$ and to the other implication I have to prove that $G/\langle g_1,...,g_{n-1}\rangle$ is abelian but I'm not sure
You are right about both directions, you just have to carry the arguments out a bit further.
For the first direction, if $[G,G] < \langle g_1,...,g_{n-1}\rangle$ then it is indeed true, as you suspected, that for each $i=1,...,n-1$ and each $g \in G$ we have $$g g_i g^{-1} \in \langle g_1,...,g_{n-1}\rangle $$ because $g g_i g^{-1} g_i^{-1} \in [G,G] \in \langle g_1,...,g_{n-1}\rangle$, and also $g_i^{-1} \in \langle g_1,...,g_{n-1}\rangle$, and therefore $g g_i g^{-1} g_i^{-1} = (g g_1 g^{-1}) g_i \in \langle g_1,...,g_{n-1}\rangle$. And this implies that $\langle g_1,...,g_{n-1}\rangle$ is normal: for example $$g (g_1 g_2^{-1} g_3) g^{-1} = (g g_1 g^{-1})(g g_2^{-1} g^{-1}) (g g_3 g^{-1}) \in \langle g_1,...,g_{n-1}\rangle $$ and similarly for any word in the generators of $\langle g_1,...,g_{n-1}\rangle$
You're right about the other implication too: in fact $G / \langle g_1,...,g_{n-1} \rangle$ is abelian because it is generated by a single element, namely the image of $g_n$. Therefore $G / \langle g_1,...,g_{n-1} \rangle$ is cyclic, hence abelian.