Prove $m\mid \gcd(a_1,\dots,a_k)\text{ implies }\gcd(a_1,\dots,a_k)= m\gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})$ 
Let $m,a_1,\dots,a_k\in\mathbb{N},$ show that:
  $$m\mid \gcd(a_1,\dots,a_k)\text{ implies } \gcd(a_1,\dots,a_k)= m\gcd(\frac{a_1}{m}, \dots,\frac{a_k}{m}), \gcd\left(\frac{a_1}{\gcd(a_1,\dots,a_k)},\dots,\frac{a_k}{\gcd(a_1,\dots,a_k)}\right)=1$$

My Attempts:
Since $m\mid \gcd(a_1,\dots,a_k)$ and $\gcd(a_1,\dots,a_k)\mid a_j$ where $j\in[1,k]\cap\mathbb{N}$ that $m\mid a_j$ we have:
\begin{align}
\frac{\gcd(a_1,\dots,a_k)}{m}\mid \frac{a_j}{m}
&\equiv \frac{\gcd(a_1,\dots,a_k)}{m}\mid \frac{a_1}{m},\cdots,\frac{\gcd(a_1,\dots,a_k)}{m}\mid\frac{a_k}{m}\\
&\Rightarrow \frac{\gcd(a_1,\dots,a_k)}{m}\le \gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})\\
&\Rightarrow \gcd(a_1,\dots,a_k)\le m\gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})\tag{1}
\end{align}
Similarly we have:
\begin{align}
\gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})\mid \frac{a_j}{m}
&\equiv \gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})\mid \frac{a_1}{m},\cdots,\gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})\mid\frac{a_k}{m}\\
&\equiv m\gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})\mid a_1,\cdots,m\gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})\mid a_k\\
&\Rightarrow m\gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})\le\gcd(a_1,\cdots,a_k)\\
&\Rightarrow m\gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})\le\gcd(a_1,\dots,a_k)\tag{2}
\end{align}
Put $(1)$ and $(2)$ together that:
$$m\gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})\le\gcd(a_1,\dots,a_k)\le m\gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})$$
That $\gcd(a_1,\dots,a_k)= m\gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})$
Let $m=\gcd(a_1,\dots,a_k)$ we have:
$$\gcd(a_1,\dots,a_k)\gcd(\frac{a_1}{\gcd(a_1,\dots,a_k)},\dots,\frac{a_k}{\gcd(a_1,\dots,a_k)})=\gcd(a_1,\dots,a_k)$$
Divide both side by $\gcd(a_1,\dots,a_k)$, proved $\gcd\left(\frac{a_1}{\gcd(a_1,\dots,a_k)},\dots,\frac{a_k}{\gcd(a_1,\dots,a_k)}\right)=1\tag*{$\square$}$
Is my proof correct, and are there better approaches $?$
Thanks for your help.
 A: Using inequalities seems like overkill.
As a general property, we have and will use $$m\mid \gcd(a_1,\dots,a_k) \hspace{0.2cm}\Leftrightarrow \hspace{0.2cm} m \mid a_{p}, \forall p\in\{1..k\} \hspace{2cm} (1)$$
Assume $\gcd\left(\frac{a_1}{\gcd(a_1,\dots,a_k)},\dots,\frac{a_k}{\gcd(a_1,\dots,a_k)}\right)=n$ . 
Then by $(1)$ we get $n \mid \frac{a_p}{\gcd(a_1,\dots,a_k)}, \forall p\in\{1..k\}$. 
This can be written as $(n \gcd(a_1,\dots,a_k))\mid a_{p},\forall p\in\{1..k\}$
By $(1)$ we get $(n \gcd(a_1,\dots,a_k))|  \gcd(a_1,\dots,a_k)$ .
From this we get $n\in \{-1,+1\}$, but $n$ is pozitive so $n=1$. 
A: This implication does not make sense, because $m$ is never used. But the implication is always true. Suppose:
$$\gcd\left(\frac{a_1}{\gcd(a_1,\dots,a_k)},\dots,\frac{a_k}{\gcd(a_1,\dots,a_k)}\right)=n$$
This would mean, that $\frac{a_i}{\gcd(a_1,\dots,a_k)}$ is always divisible by $n$ - contraddiction, because in denominator is $\gcd(a_1,\dots,a_k)$, so there cannot be any common divisors.
