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

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

I know that the normalizer of any proper subgroup of a nilpotent group contains this subgroup properly. So I think maybe I can prove the normality of the Sylow subgroup of $G$, say $P$, by showing that $N(P)=N(N(P))$, thus showing that $N(P)=G$. (Here, $N(P)$ represents the normalizer of $P$ in $G$.) But I don't know how to complete this step.
I'll appreciate your help. Many thanks.
 A: 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.
A: 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.
A: First we need this Lemma 1:

Let G be a ﬁnite group and let $P$ be a Sylow p-subgroup of $G$ for some prime $p$, then $N_G(N_G(P)) = N_G(P)$.

The proof is here.
And we also need this Proposition 2:

Let $G$ be a nilpotent group. Then every proper subgroup of $G$ is properly contained in its normalizer, that is $H < N_G(H)$ whenever $H < G$.

( Post question and comment me for the proof. )
Let $G$ be nilpotent and $P$ be a Sylow subgroup of $G$ for some prime p. Let $H = N_G(P)$. Combining both Lemma 1 and Proposition 2, we have:

by Lemma 1, $N_G(H) = H$;
by Proposition 2, $H < N_G(H)$ whenever $H < G$.

So $H$ must be equal to $G$. Thus $N_G(P) = H = G$ or $P \unlhd G$.
