Proof for formula for number of divisors

Question

For any natural number $n>1$, with factorization $$ n = q_1^{\alpha_1}q_2^{\alpha_2}q_3^{\alpha_3}....q_m^{\alpha_m} $$ we can find the number of divisors by using the formula, $$ (\alpha_1+1)(\alpha_2+1)....(\alpha_m+1) $$ How do I find this formula. Is there a proof for it?

Answer 1

To prove this first consider the number of the form $n=p^{\alpha}$. The divisors are $1,p,p^2,\cdots, p^{\alpha}$, i.e. $d(p^{\alpha})=\alpha+1$. Now, consider $n=p^{\alpha}q^{\beta}$, where $p,q$ are co-prime. The divisors would be : $$1,p,p^2,\cdots p^{\alpha}$$

$$q,pq,p^2q,\cdots,p^{\alpha}q$$ Similarly, upto: $$q^{\beta},pq^{\beta},\cdots,p^{\alpha}q^{\beta}$$ In this case, $d(p^{\alpha}q^{\beta})=(\alpha+1)(\beta+1)$. Hence $d(n)$ is multiplicative function.

For any natural number $n>1$, with prime factorization , where $q_1,q_2\cdots q_m$ are distinct primes and $m\ge1 $ $$ n = q_1^{\alpha_1}q_2^{\alpha_2}q_3^{\alpha_3}....q_m^{\alpha_m} $$ we can find the number of divisors by using the formula, $$ (\alpha_1+1)(\alpha_2+1)....(\alpha_m+1) $$