Inductive proof for the sums of exponents of prime factors?

Question

I saw this somewhere and remembered it when presented with a question concerning the number of positive integers that are divisors for some integer.

Consider :

For some prime factorization: $$a^{i}*b^{j}*c^{k}*...z^{n}=d$$ $d \in\mathbb{Z}$ and $i,j,k...n \in\mathbb{N_0}$

then $(i+1)(j+1)(k+1)(n+1)$ yields the number of all positive divisors of $d$.

Is there a proof by induction for this? Is this an already existing theorem or lemma that I might have missed?

Answer 1

Yeah, you can do induction on the number of distinct prime divisors.

If $n=p^k$ then the divisors of $n$ are $1, p, \dots, p^k$, so there are $k+1$ of them.

$If n=p_1^{k_1} \dots p_{r+1}^{k_{r+1}}$ then by induction hypothesis, $n$ has $(k_1+1) \dots (k_r + 1)$ divisors $d$ coprime with $p_{r+1}$, and for each of these $d$ you get exactly $k_{r+1}+1$ distinct divisors of $n$, namely $d, p_{r+1}d, \dots, p_{r+1}^{k_{r+1}}$.