Consider $n, k ∈ \mathbb{N}$. Prove that $\gcd(n, n + k)\mid k$ 

Consider $n, k\in\mathbb N$. Prove that $\gcd(n, n + k)\mid k$


Here is my proof. Is it correct?
A proposition in number theory states the following:

$$\forall a, b \in \mathbb{Z}, \ \gcd(n,m)\mid an+bm$$

A corollary (consequence) to this is that if $an+bm=1$ for some $a, b \in \mathbb{Z}$, then $\gcd(n,m)=1$. 
Again in our problem $(n+k)-n=k$, therefore we can use this to see that $\gcd(n,n+k)=k$.

Proof: Let $m=gcd(n,n+k)$ for $n \in \mathbb{Z}$ and $k \in \mathbb{N}$. Then $m \mid n$ and $m \mid (n+k)$ and so $n=mx$ and $(n+k)=my$ for $x,y \in \mathbb{Z}$. Observe that $(n+k)-n=k=my-mx=m(y-x)$. Since $y-x$ is an integer, $m \mid k$. Thus, $gcd(n,n+k) \mid k$.
