Finitely generated nilpotent group where every element is of finite order is finite

Hint: first induction step: $G/[G,G]$ is an abelian group that inherits every element having finite order and being finitely generated. Hence this group must be finite. Now look at $\gamma_2(G)/\gamma_3(G)$. This group is also abelian and has finite index (because of the previous step) in $G/\gamma_3(G)$. Since also $G/\gamma_3(G)$ is finitely generated, $\gamma_2(G)/\gamma_3(G)$ must be finitely generated. Again $\gamma_2(G)/\gamma_3(G)$ is a finitely generated abelian group with all elements of finite order, whence finite. Can you finish?

This is a nice problem because you don't use induction directly but prove it in several steps (only one of which requires induction):

Step 1: If $N\trianglelefteq G$ such that $[G,N]\le Z(G)$ and both $N$ and $G$ are generated by a finite number of elements of finite order, then $[G,N]$ is also generated by a finite number of elements of finite order.

(This can be proved by commutator arithmetic.)

Step 2: For $G$ as above, all elements of the lower central series are generated by finitely many elements of finite order.

(That's the part that requires induction, but is otherwise a trivial consequence of step 1.)

Step 3: An abelian group generated by a finite number of elements of finite order is finite.

(This is trivial as well.)

Putting the three steps together solves your problem.