Classical intro to modern Number theory

Question

I'm self-studying Classical Intro to Modern Number Theory, by Kenneth Ireland and Michael Rosen, and I am stuck on a simple proof on page $34$:

Suppose $a_1, a_2 ...,a_t$ all divide $n$, and that $gcd(a_i, a_j) = 1$ for $i \neq j$. then $a_1\cdot\ a_2\cdot \ldots \cdot a_t$ all divide $n$.

The book proves by induction:

$a_1\cdot a_2\cdot \dots a_{t-1}$ divide $n$. Then $gcd(a_t, a_1\cdot \ldots a_{t-1}) = 1$. Then $\exists$ $r, s$ such that $r\cdot a_t + s\cdot a_1\cdot \ldots a_t = 1$. Multiply both sides by $n$. Inspection shows that the left-hand side is divisible by $a_1\cdot a_2\ldots \cdot a_t$ and the result follows.

I don't understand the multiply by $n$ and inspection part. It seems straight forward but I'm spacing.

Answer 1

Isn't it much easier just using prime factor decomposition, writing $n=\prod p_i^{k_i}$ and noticing that by definition, as $gcd(a_i,a_j)=1$ and $a_i\mid n$ each $a_i$ is a subproduct of the $p_i's$ that make up $n$ and all subproducts use distinct sublist of $p_i's$?

Answer 2

We have $n\cdot r\cdot a_t +n \cdot s\cdot a_1\cdots a_{t-1} = n$.

Since $a_t$ divides $n$ by hypothesis, $ a_1\cdots a_t$ divides $n\cdot s\cdot a_1\cdots a_{t-1}$. And $a_1\cdots a_{t-1}$ divides $n$ by induction hypothesis,thus $ a_1\cdots a_t $ divides $n\cdot r\cdot a_t$.

Therefore $ a_1\cdots a_t $ divides the sum $n\cdot r\cdot a_t +n \cdot s\cdot a_1\cdots a_{t-1} = n$.