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

We want to show that $\gcd (a, a + k) | k$
Let $\gcd (a, a + k) = d$ say, where $d$ is an integer
$d| a     \implies  a     = md$    , $m$ is an integer
$d| a + k  \implies  a + k = nd$ , $n$ is an integer
$a + k - a = nd - md $
$k = d (n - m)$
$\implies d | k$
$\implies \gcd(a,a + k) | k$    [as required]
