I need to do a proof by induction without the use of a closed form (I do know it's possible to find one, but we aren't supposed to do it that way).
Here's the prompt:
Let $p_1, p_2, p_3, \ldots, p_m$ be distinct primes. Let $a_1, a_2, a_3, \ldots, a_m$ be positive integers. Suppose the following
\begin{equation} N = (p_{1})^{a_{1}} \cdot (p_{2})^{a_{2}} \cdot (p_{3})^{a_{3}} \cdot \ldots \cdot (p_{m})^{a_{m}} \end{equation}
How many divisors does $N$ have?
I already figured out that the number of divisors of N is:
$Divisors\hspace{.2cm} of\hspace{.2cm} N = (a_{1} + 1) \cdot (a_{2} + 1) \cdot (a_{3} + 1) \cdot \ldots \cdot (a_{m} + 1) $
I asked my professor how to do an induction proof with only this, and he said:
"To interpret this as an induction proof, do induction on m. So, assuming that some number $N$ has m prime factors, consider a number $N'$ that has $m+1$ prime factors. Show that if N has $(a_1 + 1)\cdot(a_2+1)\cdot(a_3+1)\cdot \ldots \cdot(a_m+1)$ total factors, then $N'$ has $(a_1 + 1)\cdot(a_2+1)\cdot(a_3+1)\cdot \ldots \cdot(a_m+1)\cdot(a_{m+1} + 1)$ total factors."
I am still struggling to understand how to do this. Of course you could say, let:
$N = p_{1}^{a_{1}}p_{2}^{a_{2}}$
and let :
$N' = p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}}$
and then :
$N' = N \cdot p_{3}^{a_{3}}$
but I really don't understand how to proceed from there in order to reach any sort of conclusion.
Any help would be hugely appreciated. Thanks.