We say $g ∈ N = gcd(a, b)$ if these are true:
$1.$ $g$ divides both $a$ and $b$.
$2.$ Whenever $d ∈ N$ divides $a$ and $b$, then $d$ also divides $g$.
($N$ is the set of natural numbers and $gcd$ is the greatest common denominator).
So, to show $n = gcd(m, n)$, I need to satisfy $1.$ and $2.$
$1.$ $n|m$ is given, and $n|n$ is true because $n$ can be expressed as $n = kn$ for $k = 1$.
$2.$ Suppose $d ∈ N$, and $d|m$ and $d|n$. Then $d|n$ is given.
Therefore $gcd(m, n) = n$ when $n|m$.
Is this a correct proof? I was unsure of it because so much was already given to me without much actual work having to be done.