Show that if $G$ is a finite nilpotent group, then every Sylow subgroup is normal in $G$

A fairly simple way to do this is to note that a normal Sylow-subgroup is always characteristic. Thus, since you can make a subnormal chain from the Sylow-subgroup to the group itself, this Sylow-subgroup must in fact be normal in each subgroup in the chain and thus in the entire group.

Let $G$ be any finite group, and let $P$ be a Sylow $p$-subgroup of $G$, and $N(P)$ the normalizer in $G$ of $P$.

Note that $P$ is a Sylow $p$-subgroup of $N(P)$, and in fact is normal in $N(P)$; that means that $P$ is the only Sylow $p$-subgroup of $G$ that is contained in $N(P)$.

Now, suppose $g\in G$ normalizes $N(P)$ (that is, $g\in N(N(P))$). Then $g^{-1}N(P)g = N(P)$. We also know that $g^{-1}Pg$ is a Sylow $p$-subgroup of $G$, and since $P\subseteq N(P)$, then $g^{-1}Pg\subseteq g^{-1}N(P)g = N(P)$.

What does that tell you about $g^{-1}Pg$? Conclusion?

Note that this particular result does not require you to assume $G$ is nilpotent.