Number of nilpotent groups equals product of number of groups of order each maximal prime power divisor

Statement
Suppose $$n$$ is a natural number with prime factorization:

$$n = p_1^{k_1}p_2^{k_2} \dots p_r^{k_r}$$

where $$p_1,p_2,\dots,p_r$$ are distinct primes and $$k_1,k_2,\dots,k_r$$ are natural numbers.

Then the number of nilpotent groups of order $$n$$ is the product of the number of groups of order $$p_i^{k_i}$$ for all $$1 \le i \le r$$:

(Number of groups of order $$p_1^{k_1}$$) $$\times$$ (Number of groups of order $$p_2^{k_2}$$) $$ \times$$ $$\dots$$ $$\times$$ (Number of groups of order $$p_r^{k_r}$$)

Facts used

 * 1) uses::Equivalence of definitions of finite nilpotent group