Let $p_1, p_2, p_3, \cdots, p_m$ be distinct prime numbers and let $a_1, a_2, a_3, \cdots, a_m$ be positive integers. If $N = p_1^{a_1}\cdot p_2^{a_2}\cdot p_3^{a_3}\cdot\cdots\cdot p_m^{a_m}$, how many divisors does $N$ have? To prove a formula for this, you should use the multiplication rule and an induction proof.
So I found that for some prime number $p$ to some power $a$, there are $a + 1$ divisors. Example: $3^3 = 27$, divisors = $1, 3, 9, 27$, i.e., it has four divisors $(3 + 1)$.
For the inductive step, I have "assume $p_1^{a_1}\cdot p_2^{a_2}$ has $(a_1 + 1)\cdot(a_2 + 1)$ divisors. How should I include "a prime number to some power has that power $+ 1$ divisors"? Feeling stuck atm.
Wow. Not sure induction is necessary.
I'd say as $n= p_1^{a_1}p_2^{a_2}....p_m^{a_m}$ has only $p_1, ... p_m$ as prime factors all factors must be of the form $p_1^{k_1}.... p_m^{k_m}$ (where $k_i$ can be $0$) and that as $a_i$ is the highest power that of $p_i$ that can divide $n$ then each $k_i$ can be as little as $0$ and as high as $a_i$. There are $a_i+1$ options and so it is basic combinatorics that there are $(a_1 + 1).... (a_m+1)$ such factors.
That ought to be a good proof.
But if you want to (or are asked to do it) by induction:
Base case: The factors of $p^a$ are $1, p, p^2.... p^a$ and there are $a+1$ such factors.
Pf: $p$ is prime is indivisible so the only prime factors of $p^a$ is $p$ is every factor is a power of $p$. For any $k > a$, $p^k\not \mid p^a$ and for every $0\le k \le a$ then $p^k*p^{a-k} = p^a$ so all $p^k$ are factors and there are $a+1$ of them.
Inductive case: If $N= p_1^{a_1}.... p_m^{a_m}$ has $(a_1 + 1)....(a_m+1)$ factors and for $M=p_{m+1}^{a_{m+1}}$ has $a_{m+1}+1$ factors (as per base case), than $P=NM =p_1^{a_1}.... p_m^{a_m}p_{m+1}^{a_{m+1}}$ will have $(a_1 + 1)....(a_m+1)(a_{m+1}+1)$ factors.
Pf: $N$ and $M$ are relatively prime. Let $f$ be a factor of $NM$. Let $d=\gcd(N,f)$ and $c=\gcd(M,f)$. As $N$ and $M$ are relatively prime $f = dc$. So the only factors of $NM$ will be of the form $dc$ where $d|N$ and $c|M$ and every number of that form will be a factor.
As there are $(a_1 + 1)....(a_m+1)$ such $d$ and $a_{m+1} + 1$ such c. There are $(a_1 + 1)....(a_m+1)(a_{m+1}+1)$ such $cd$.