A very easy way to prove that $G'\neq G$ is to note that $G'$ is the smallest normal subgroup of $G$ such that $G/G'$ is abelian. If you can find any proper normal subgroup of $G$, $N$, such that $G/N$ is abelian, then it will follow that $G'\subseteq N$, and so $G'\neq G$.
For example: have you proven that a maximal subgroup of a $p$-group must be normal? This can be done easily by induction, using the fact that $Z(G)$ (the center) is nontrivial. With that in hand, it follows that $G'$ is always contained in a maximal subgroup of $G$, and you'd be done.
(The previous exercise you quote can't help you because it assumes your group is nilpotent, and here you are trying to prove that a group is nilpotent).
Added. Another way of proving this:
Show that if $f\colon G\to K$ is an onto group homomorphism, then $f([G,G])=[K,K]$. In particular, if $[K,K]\neq K$, then $[G,G]\neq G$. Then use the fact that if $G$ is a $p$-group then $Z(G)\neq \{1\}$ and induction to show that either $Z(G)=G$ (so $[G,G]=\{1\}$) or else $[G/Z(G),G/Z(G)]\neq G/Z(G)$ and get the result.
Note: Your notation for the lower central series is uncommon. It is more common to use $G_1=G$ and $G_{n+1}=[G_n,G](=[G,G_n])$.
Added. I wasn't paying enough attention when I answered this, and what Derek Holt points out in comments is quite true: it is not enough to prove that $[G,G]$ is nilpotent in order to prove that $G$ is nilpotent, even if $[G,G]\neq G$: there are groups where $[G,G]$ is nilpotent, but $G$ is not. For example, $S_3$ has $[G,G]\cong A_3$, which is abelian, but $S_3$ is not nilpotent.
Instead, we can proceed by induction on $n$, where $|G|=p^n$.
If $n=1$ or $n=2$, then $G$ is abelian, hence $G$ is nilpotent.
Assume that $G$ has order $p^n$. If $Z(G)=G$, then $G$ is abelian and hence is nilpotent. Otherwise, consider $\mathfrak{G}=G/Z(G)$. This is a $p$-group of order strictly less than $p^n$ (since the center of a finite $p$-group is always nontrivial), so by the induction hypothesis, $\mathfrak{G}$ is nilpotent. Now we use the lemma mentioned above, slightly strengethened (I will use $H_n$ to denote the $n$th term of the lower central series of $H$, that is, $H_n$ is what the original post calls $C^n(H)$):
Lemma. Let $f\colon G\to K$ be a group homomorphism. Then for every $m$ we have $f(G_m) \subseteq K_m$. If $f$ is onto, then $f(G_m)=K_m$ for all $m$.
Proof. The result holds for $m=1$. Assume $f(G_m)\subseteq K_m$. A generator for $G_{m+1}$ is of the form $[g,g']$, where $g'\in G_m$, hence $f([g,g']) = [f(g),f(g')]\in [K,K_m]=K_{m+1}$, so $f(G_{m+1})\subseteq K_{m+1}$, as claimed. Assume now that $f$ is onto and $f(G_m)=K_m$. A generator for $K_{m+1}$ is of the form $[k,k']$ with $k\in K$ and $k'\in K_{m+1}$; since $f$ is onto there exists $g\in G$ with $f(g)=k$; since $f(G_m)=K_m$, there exists $g'\in G_m$ with $f(g')=k'$. Hence $f([g,g'])=[f(g),f(g')] = [k,k']$, and since $[g,g']\in G_{m+1}$, it follows that $K_{m+1}\subseteq f(G_{m+1})$, as desired. $\Box$
Now, since $\mathfrak{G}$ is nilpotent, there exists $r$ such that $\mathfrak{G}_r$ is trivial; therefore, by the lemma, $f(G_r)$ is trivial, so $G_rZ(G) = Z(G)$. Hence $G_r\subseteq Z(G)$, so $G_{r+1}=[G,G_r]\subseteq [G,Z(G)] = \{1\}$, so $G$ is nilpotent, as claimed. $\Box$
Best Answer
What group do you suppose is nilpotent, so that you can apply this to it? It's not like you know $G$ is nilpotent yet: that's what you're trying to prove.
Suppose you believe the hint. Then in the case that $N_G(P)\neq G$, you argue that it's contained in a maximal subgroup of $G$, which is equal to its own normalizer. But that contradicts the hypotheses...