Exercise 20.12 of Jech's Set Theory says:
Let $\kappa$ be the least huge cardinal and let $\mu$ be the least supercompact cardinal. Then $\kappa < \mu$.
The hint below says:
If $\kappa = \mu$ then by 20.27, 20.25, 20.24 and 20.23 we get: $V_\mu \models \text{VP}$, $V_\mu \models (\text{$\exists$ supercompact $\alpha$})$, there is a supercompact $\alpha < \mu$, a contradiction. If $\kappa < \mu$, let $j : V \to M$ with $\lambda = j(\kappa)$ and $M^\lambda \subseteq M$. Since $\mu$ is supercompact, let $i : V \to N$ be such that $i(\mu) > \lambda$ and $\color{red}{V_{\lambda+2} \subseteq N}$. If $U$ is a normal measure witnessing the hugeness of $\kappa$, then $U \in N$, and hence $N \models (\text{$\exists$ huge cardinal below $i(\mu)$})$. Thus there exists a huge cardinal below $\mu$, a contradiction.
Note that there is probably a typo - the second statement should start with "If $\mu < \kappa$...".
I'm confused with the red part $V_{\lambda+2} \subseteq N$. How can we guarantee that there exists an embedding $i : V \to N$ such that $N$ contains $V_{\lambda+2}$?
Since $\mu$ is supercompact, we can find $\delta\geq|V_{\lambda+2}|$ such that $N$ is closed under $\delta$-sequences.
In particular, $V_{\lambda+2}\in N$, since we can encode it as a sequence of ordinals and then apply Mostowski's collapse lemma. But if $V_\alpha\in N$, then $V_\alpha^N=V_\alpha$. In particular for $\alpha=\lambda+2$.