Reason for order of multiplication of ordinals I understand that addition and multiplication on ordinal numbers cannot be commutative and I know why $1+\omega$ and $2\times\omega$ must be different from $\omega+1$ and $\omega\times2$, respectively. For addition, I see why $1+\omega$ is just $\omega$, because it represents a position after $1$ and then $\aleph_0$ elements, and I can include the first element into the rest and the position wouldn't change.
What I am intrigued about is why $\omega+\omega$ is equal to $\omega\times2$ and not $2\times\omega$, as I would think.
How I interpret an expression like $5×2$ is "five-times two", like two seen five-times, as in $2+2+2+2+2$. In the context of ordinals, I would also imagine the expression $\omega+\omega$ (as I see omega two-times) to be $2\times\omega$. In contrast, I would interpret an expression like $\omega\times2$ to be $\underbrace{2+2+2+...}_\omega$, which also corresponds to a set of $\aleph_0$ elements and thus be equal to $\omega$. However, the usual definition is the exact opposite.
I understand that notation is not the underlying mathematics and one may freely redefine the operator in terms of changing the order of operands, but this definition seems to me arbitrary and inconsistent. Is there any particular reason why it was defined so?

Personally, I accepted this order of multiplications by noting that $\omega^{(\omega + 1)} = \omega^\omega \cdot \omega.$
More generally, $\alpha^{(\beta + \gamma)} = \alpha^\beta \cdot \alpha^\gamma$.
