Left subtraction and left division with remainder for ordinals is unique From Wikipedia:

Furthermore, one can define left subtraction for ordinals $\beta \leq \alpha$: there is a unique $\gamma$ such that $\alpha = \beta + \gamma$.

I understand both the definitions of ordinal addition as stated on Wikipedia. However, is there any simple argument to see that the quoted statement is true? The first definition using disjoint union of well-ordered sets is more preferable though, rather than the definition by induction.
I roughly think that as $\beta$ is a subset of $\alpha$, taking its union with the well-ordered set $\gamma  = \alpha \setminus \beta$ works, but I'm not sure about uniqueness. Is it just because set complements are unique?
More generally, it seems the following holds:

Left division with remainder: for all $\alpha$ and $\beta$, if $\beta > 0$, then there are unique $\gamma$ and $\delta$ such that $\alpha = \beta·\gamma + \delta$ and $\delta < \beta$.

Is there a simple approach to prove this too?
 A: Yes, since $\alpha\setminus\beta$ is well-ordered, it is order-isomorphic to a unique ordinal $\gamma$, and the definition of ordinal addition using the disjoint union of $\beta$ and $\gamma$ shows that $\beta+\gamma=\alpha$. If $\beta+\delta=\alpha$, you can use that same definition to see that $\delta$ is also order-isomorphic to $\alpha\setminus\beta$, so it must be $\gamma$. (I am assuming the result that a well-ordered set is order-isomorphic to a unique ordinal, which follows from the result that distinct ordinals are not order-isomorphic.)
For the second question, there is certainly a least ordinal $\mu$ such that $\beta\cdot\mu>\alpha$. If $\mu$ were a limit ordinal, we’d have $\beta\cdot\xi\le\alpha$ for each $\xi<\mu$ and hence $\beta\cdot\mu\le\alpha$, so $\mu$ must be a successor, say $\mu=\gamma+1$. Then $\beta\cdot\gamma\le\alpha$, and by the first result there is a unique $\delta$ such that $\alpha=\beta\cdot\gamma+\delta$. And $\beta\cdot\gamma+\beta=\beta\cdot(\gamma+1)>\alpha$, so we must have $\delta<\beta$.
