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 $\gcd(n,n+k)=k$.

Even easier $d=\gcd(n,n+k)$


*

*if $d=1$ then $d \mid k$

*if $d>1$ then $d \mid n$ and $d \mid n+k$ or $d\cdot q_1=n$ and $d\cdot q_2=n+k$. Altogether $d\cdot q_2 = d\cdot q_1 +k$ or $d(q_1-q_2)=k \Rightarrow d \mid k$

