Let $G$ be a nilpotent group generated by a finite set of torsion (i.e. finite order) elements. Show that $G$ is finite. Given: Let $G$ be a nilpotent group generated by a finite set of torsion (i.e. finite order) elements. 
Show that $G$ is finite.
Also would love to know if it's possible to show that an infinite finitely generated nilpotent group has an infinite center.
 A: The following argument is incomplete, although I did not realize it until after writing it out. In fact, I now see that it is a fleshing out of the comment by runway44 above. After further thought, the missing step (torsion generation of $[G,G]$) is quite substantial and indicates this is not the right approach. Instead, the argument outlined by Derek Holt appears to be the correct path.

We prove the result by induction on the length of the lower central series. The base case consists of an abelian group, in which case the claim is immediate.
The inductive hypothesis applied to $[G,G]$ implies that it is a finite group. Let $g_1,\ldots,g_n$ be a generating set of torsion elements for $G$. Then every $g\in G$ has a representation of the form
$$
g=h\prod_{i=1}^{n}g_i^{x_i},\qquad h\in [G,G],
$$
where each $x_i$ ranges from $0$ to $\textrm{ord}(g_i)-1$. Indeed, we may iteratively all instances of $g_1$ to the end (keeping a commutator at the start of $g$), then $g_2$, and so on, always keeping track of the commutators incurred during the rearrangement in $h$.
From this representation, we obtain that $|G|\leq \bigl|[G,G]\bigr| \prod_{i=1}^n\textrm{ord}(g_i)$, in particular showing that $G$ is finite.
