set theory proof explanation

Question

The following lemma is taken from the book 'Introduction to Set Theory' by Hrbacek and Jech. chapter $6$ normal form

Can anyone explain to me why the first sentence holds ( the existence of $\delta$)?

Answer 1

It follows from the multiplication being monotone, $\alpha\geq 1$ implies $\alpha\cdot (\gamma+1)\geq 1\cdot (\gamma+1)$

Answer 2

$\delta=\gamma+1$ also I think $\alpha \geq 1$

Answer 3

{$\beta$ $\in$ Ord|$\alpha\beta>\gamma$$\land\beta\le\gamma+1$} is non empty,it has a least elemant $\beta_{0}$

Assume $\beta_{0}$ is a limit ordinal ,then $\alpha\beta_{0}=sup${$\alpha\beta^{'}_{0}$|$\beta^{'}_{0}<\beta_{0}$}

but since $\forall\beta^{'}_{0}<\beta_{0},\alpha\beta^{'}_{0}\le\gamma$

$\alpha\beta_{0}$=$sup${$\alpha\beta^{'}_{0}$|$\beta^{'}_{0}<\beta_{0}$}$\le\gamma$ a controdiction.