G torsion-free, $H \leq G$ and $[G:H] < \infty$ imply equal nilpotent classes Trying to prove:

If $G$ is a torsion-free nilpotent group and $H \leq G$ has finite index, then the nilpotence classes of $G$ and $H$ are equal.

Obviously the class of H is at most that of G by taking intersections. I also know every factor group of upper central series for G is a torsion-free abelian group, and it seems intuitively sensible that H having finite index means it somehow preserves the "infiniteness" from the larger group. 
I feel like there's some small bridge I'm missing between the index being finite and, maybe, the normal series of intersections of H with the central series of G not having any trivial factors. I'd appreciate a small hint to which direction to look or just a statement of what theorem I should be using; I have the nagging feeling it's something I should already know well and am just forgetting.
 A: $\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$This is a very slight variation on the suggestion of Derek Holt.
Consider the core $K$ of $H$, that is, the intersection of all conjugates of $H$. This has also finite index in $G$, and it is normal in $G$. We will prove that $K$ has the same nilpotence class as $G$, which implies the result for $H$.
Let us prove that for all $i$ we have that $\gamma_{i}(G)/\gamma_{i}(K) \gamma_{i+1}(G)$ has finite exponent. This is true by assumption if $i = 1$, as $G/K$ is finite, so take $m_{1} = \Size{G/K}$. Suppose  $\gamma_{i}(G)/\gamma_{i}(K) \gamma_{i+1}(G)$ has finite exponent $m_{i}$, let $g \in \gamma_{i}(G)$, $h \in G$. Then
$$
[g, h]^{m_{i} m_{1}} \in [g^{m_{i}}, h^{m_{1}}] \gamma_{i+2}(G) \le  [\gamma_{i}(K), K] \gamma_{i+2}(G) = \gamma_{i+1}(K) \gamma_{i+2}(G). 
$$
So if $G$ has class $n$ and $K$ has class less than $n$, we have that
$$
\gamma_{n-1}(G)/\gamma_{n}(K) \gamma_{n+1}(G) \cong \gamma_{n-1}(G)
$$
has finite exponent, and thus is trivial, a contradiction.
