Exponent of a finite group equals product of exponents of its Sylow subgroups

Statement
 Suppose $$G$$ is a finite group and $$\{ p_1,p_2, \dots, p_r \}$$ is the set of prime numbers dividing the order of $$G$$. For each $$p_i$$, let $$P_i$$ be a $$p_i$$-Sylow subgroup of $$G$$. Then, we have:

$$\mbox{Exponent of } G = \prod_{i=1}^r (\mbox{exponent of } P_i)$$

Related facts

 * Exponent of a finite group has precisely the same prime factors as order
 * Exponent divides order