Show that $\forall_{n,m,k\in \mathbb{N}}(k|nm$ and gcd$(n,k)=1 \Rightarrow k|m)$

Question

Please help me

Definitons

gcd := greatest common divisor

Let $k,n\in \mathbb{Z}$ One says that $k$ divides $n$, in symbols $k|n$ if there exists a $l \in \mathbb{Z}$ s.t. $n=kl$

For two natural numbers $m,n \in \mathbb{N}$ the greatest common divisior is defined (symblolic gcd$(k,n)$), the biggest natural number $k$ with $k|m$ and $k|n$

Answer 1

If $m = 0$ the implication is obvious

If $m \ne 0$
Then we calculate $gcd(nm,km) = m*gcd(n,k) = m$ becquse $gcd(n,k) = 1$
We know thqt $k|nm$ and $k|km$ thus $k|gcd(nm,km)$
And $gcd(mn,mk) = m$
$$\implies k|m$$