Why is conductor of a Dirichlet character the product of conductors of other Dirichlet characters? Let $n=\prod_{i=1}^np_i^{e_i}$ with $p_i$ different prime numbers and $e_i$ positive integers. Given a Dirichlet character $\chi$ modulo $n$ we can define the characters $\chi_i$ (modulo $p_i^{e_i}$) by $\chi_i(a):=\chi(a_i)$ where $a_i$ is such that $a_i\equiv a \mod p_i^{e_i}$ and $a_i\equiv 1\mod p_j^{e_j}$ for $j\neq i$ (by using CRT). Using these characters one can factor $\chi$ uniquely as $\chi=\chi_1\ldots\chi_r$. Let $c_{\phi}$ be the conductor of the Dirichlet character $\phi$ modulo $k$, i.e. the smallest number $d$ such that $\phi(a)=1$ for all $a\in\mathbb{Z}$ satisfying $\gcd(a,k)=1$ and $a\equiv1\mod d$. 
I want to show that $c_{\chi}=c_{\chi_1}\ldots c_{\chi_r}$ and I can prove that $c_{\chi}\mid c_{\chi_1}\ldots c_{\chi_r}$. So it only remains to show that we cannot have $c_{\chi}<c_{\chi_1}\ldots c_{\chi_r}$. In the book "Modular Forms, a Computational Approach'' by William A. Stein there is a solution (see page 72 of https://books.google.nl/books?id=blaZAwAAQBAJ&printsec=frontcover&hl=nl#v=onepage&q&f=false) which I don't fully understand. The argument starts by showing that $c_{\chi}\mid c_{\chi_1}\ldots c_{\chi_r}$ and then proceeds by assuming that there is a prime $p$ such that $\text{ord}_{p}(c_{\chi})<\text{ord}_p(c_{\chi_j})$ for some $j$ to arrive at a contradiction. The proof says that if this is the case then we can factor $\chi$ as a product of (local) prime power characters differently (see first paragraph). Which gives the desired contradiction.
My question is why we can factor $\chi$ differently if we assume  $\text{ord}_{p}(c_{\chi})<\text{ord}_p(c_{\chi_j})$?
 A: Here is a possible answer:
First we know that
$$\operatorname{ord}_{p_j}(f(\chi))<\operatorname{ord}_{p_j}(f(\chi_j))$$
suggests that there exists some $a\equiv 1\pmod{f(\chi)}$, which implies $$a\equiv 1\pmod{p_j^{\operatorname{ord}_{p_j}(f(\chi))}},$$ 
but $a\not\equiv1\pmod{f(\chi_j)}$. This integer $a$ satisfies
    $$\chi(a)=1$$
    but
    $$\chi_j(a)\neq 1$$
    because $p_j^{\operatorname{ord}_{p_j}(f(\chi))}<f(\chi_j)$. Thus, $\chi$ cannot be factorized as $\chi=\chi_1\cdots\chi_r$ unless there exists some $\prod\chi_{j'}$ such that $\prod\chi_{j'}(a)=1/\chi_j(a)$. Since $\chi_j$ is a $k_j$th root of unity, we have $\prod\chi_{j'}(a)=e^{-2\pi ix/k_j}$ where $0<x<k_j$, which is impossible because we know $(k_j,k_j')=1$ for all $j'$ so there is no $x$ that can make $x/k_j=x'/\prod k_{j'}$ unless $x'k_j=x\prod k_{j'}$, which implies $k_j\mid x$. Thus a contradiction, which implies $\chi$ must have another factorization. Which completes the proof.

This proof is quite convoluted but I think it will give you the basic idea to approach the conclusion. (Though I doubt you still need this in 2020.)
