If $G / Z(G)$ nilpotent then G is nilpotent. 
Let $G$ be a group. If $G / Z(G)$ nilpotent, then prove that $G$ is nilpotent.

The way I have worked out this question is fairly tedious, and I would be interested if there is a more straightforward approach here.
I solved this problem by showing that if $G / Z(G)$ is of nilpotence class $k$, then $G / Z(G) = Z_{k}(G / Z(G)) = Z_{k+1}(G) / Z(G)$, the latter equality shown using a tedious induction proof. Then the conclusion readily follows.
Is there a different approach to this problem, or is there a short way to prove that $Z_{k}(G / Z(G)) = Z_{k+1}(G) / Z(G)$?
 A: Along the lines of egreg's answer, you might consider proving (and using) the more general statement:

A group $G$ is nilpotent if and only if, for every normal subgroup $H$, both $H$ and $G/H$ are nilpotent.

The argument is the one he suggested.
A: Hint: given a central series
$$
Z(G)/Z(G)\lhd H_1/Z(G)\lhd H_2/Z(G)\lhd \dots \lhd H_{k-1}/Z(G)\lhd H_k/Z(G)=G/Z(G)
$$
of $G/Z(G)$, then
$$
\{1\}\lhd Z(G)\lhd H_1\lhd H_2\lhd \dots \lhd H_{k-1}\lhd H_k=G,
$$
is a central series for $G$.
A: Using this Lemma 1:

If $\pi:G \rightarrow K$ is a surjective homomorphism, then $\pi(\gamma_i(G)) = \gamma_i(K)$ for all $i$.

( Ask and comment me for the proof. )
By definition of nilpotent group we have:
$\gamma_{x+1}(G/Z(G)) = 1$.
Let $\pi:G \rightarrow G/Z(G)$ be the homomorphism. By Lemma 1:
$\pi(\gamma_{x+1}(G)) = \gamma_{x+1}(G/Z(G)) = 1$.
But $\pi(\gamma_{x+1}(G)) = 1$ means that $\gamma_{x+1}(G) \le ker \: \pi = Z(G)$. Thus:
$\gamma_{x+2}(G) = [\gamma_{x+1}(G), G] \le [Z(G), G] = 1$ because $\gamma_{x+1}(G) \le Z(G)$ and $[Z(G), G] = 1$.
As $\gamma_{x+2}(G) = 1 = \gamma_{c+1}(G)$, $G$ is nilpotent as required.
