Consecutive Integers are Coprime/Proof 2

Proof
Let $k \in \Z: k \divides h$.

Also assume $k \divides \paren {h + 1}$.

Thus:
 * $\exists a, b \in \N: a k = h, b k = \paren {h + 1}$

Then:
 * $\paren {h + 1} - h = b k - a k$

and so:
 * $1 = \paren {b - a} k$

Since the integers form an integral domain, $\paren {b - a} \in \Z$.

Thus either $k = 1$ and $b - a = 1$, or $k = -1$ and $b - a = -1$.

Therefore, only $1$ and $-1$ can be factors of both $h$ and $\paren {h + 1}$.