Order of associated Lie ring equals order of group for nilpotent group

Statement
Suppose $$G$$ is a nilpotent group. Then, the order (i.e., underlying set size) of the associated Lie ring $$L(G)$$ equals the order of $$G$$. This equality holds both in the case of finite orders and in the infinite case, with orders expressed as cardinals.

Note that we have that nilpotency class of associated Lie ring equals nilpotency class of quotient of group by nilpotent residual. In particular, this says that if the associated Lie ring is nilpotent and the group is residually nilpotent, then the group is nilpotent. Thus, the result can be applied in these cases.

Facts

 * 1) uses::Order of direct product is product of orders
 * 2) uses::Order of group is product of orders of successive quotient groups of subnormal series

Proof
Suppose $$G$$ has nilpotency class $$c$$. Denote by $$\gamma_i(G)$$ the $$i^{th}$$ member of the lower central series of $$G$$. Then, by Fact (1):

$$|L(G)| = \prod_{i=1}^c |\gamma_i(G)/\gamma_{i+1}(G)|$$

By Fact (2):

$$|G| = \prod_{i=1}^c |\gamma_i(G)/\gamma_{i+1}(G)|$$

Comparing, we obtain that $$|L(G)| = |G|$$.