Let $F(S)$ be the free group generated by the set $S$ with $|S|=n$. I need to show that there is no generating set $T$ of $F(S)$ with $|T|<n$.
So far I noticed that if $|T|$ generated $F(S)$, then we could use a surjective map $\phi:S\to T$. Due to the universal property, this extends to a homomorphism $\phi:F(S)\to F(S)$ with nontrivial kernel. Then $F(S)=\langle \phi(S)\mid\ker \phi\rangle$.
When showing the existence of free groups, one gets that $F(S)=\langle S\mid\_ \ \rangle$. But I don't know how to show that this gives a contradiction.