Asaf and Joel have answered the question. Let me add a remark that expands the fact that it helps us prove that $\kappa\times$ and $\kappa$ have the same size. All the claims here can be verified rather easily. Multiplication and exponentiation are in the ordinal sense.
It is customary to write $\Gamma(\alpha,\beta)$ for the order type of the predecessors of $(\alpha,\beta)$ under the order than Asaf denotes $\prec$. For example, $\Gamma(\omega,\omega\cdot2)=\omega^2+\omega$.
An ordinal $\alpha$ is (additively) indecomposable iff $\alpha\gt 0$ and whenever $\beta,\gamma\lt\alpha$, then $\beta+\gamma\lt \alpha$. One can easily check that the indecomposable $\alpha$ are precisely those of the form $\omega^\beta$, $\beta\gt 0$. \omega^\beta$. Say that$\alpha$is multiplicatively indecomposable iff$\alpha>0$and$\beta\gamma\lt \alpha$whenever$\beta,\gamma\lt\alpha$. Then$\alpha$is multiplicatively indecomposable iff it is$1$or has the form$\omega^{\omega^\beta}$. Ok, we can now state the remark; unfortunately I would not know who to credit for this observation, I think of it as folklore: An ordinal$\alpha$is multiplicatively indecomposable iff it is closed under Gödel pairing, that is,$\Gamma(\beta,\gamma)\lt\alpha$whenever$\beta,\gamma\lt\alpha$. In particular,$\Gamma(\kappa,\kappa)=\kappa$for any infinite cardinal$\kappa$, which of course implies that$\kappa\times\kappa$and$\kappa$have the same size. Also, if$\kappa$is uncountable, then there are$\kappa$ordinals$\alpha$below$\kappa$such that$\Gamma(\alpha,\alpha)=\alpha$. Of course, all of this works well in$\mathsf{ZF}$and all the definitions involved are absolute. I prefer a different approach when verifying that$\kappa\times\kappa$and$\kappa$have the same size, one that (again) is absolute and goes through in$\mathsf{ZF}$, but only requires the use of additively indecomposable ordinals: One first checks that there is a (recursive) bijection$h:\omega\times\omega\to\omega$with$h(0,0)=0$. Then, given ordinals$\alpha,\beta$, use their Cantor's normal form to write them as $$\alpha= \omega^{\alpha_1}n_1 + \omega^{\alpha_2}n_2 + \dots + \omega^{\alpha_k}n_k$$ and $$\beta= \omega^{\alpha_1}n'_1 + \omega^{\alpha_2}n'_2 + \dots + \omega^{\alpha_k}n'_k$$ where$\alpha_1 \gt \alpha_2 \gt \dots \gt \alpha_k$are ordinals, and$n_1,\dots,n_k, n'_1,\dots,n'_k$are natural numbers. (Note that these representations are not unique, but at least one of$n_i$and$n_i'$is non-zero iff$\alpha_i$appears as an exponent in the canonical form of$\alpha$or$\beta$). Now set $$H(\alpha,\beta)=\omega^{\alpha_1}h(n_1,n'_1)+\omega^{\alpha_2}h(n_2,n'_2)+\dots+ \omega^{\alpha_k}h(n_k,n'_k).$$ Then$H$is a bijection between$\alpha\times\alpha$and$\alpha$whenever$\alpha$is indecomposable. And an easy inductive argument, appealing to the explicit proof of Schröder-Bernstein, allows us to use$H$to argue that there is, provably in$\mathsf{ZF}$, a class function that assigns to each infinite ordinal$\alpha$a bijection between$\alpha\times\alpha$and$\alpha$. (Of course, the existence of this class function can also be argued from$\Gamma$, using that there are$\kappa$ordinals$\alpha$below$\kappa$with$\Gamma(\alpha,\alpha)=\alpha$, but this second approach is somewhat easier.) I found this argument a while ago, but then saw that Levy gives essentially the same approach in his textbook on set theory. Again, I am not sure who to credit for this construction, it seems to go back to Gerhard Hessenberg's 1906 book, "Grundbegriffe der Mengenlehre". 1 Asaf and Joel have answered the question. Let me add a remark that expands the fact that it helps us prove that$\kappa\times$and$\kappa$have the same size. All the claims here can be verified rather easily. Multiplication and exponentiation are in the ordinal sense. It is customary to write$\Gamma(\alpha,\beta)$for the order type of the predecessors of$(\alpha,\beta)$under the order than Asaf denotes$\prec$. For example,$\Gamma(\omega,\omega\cdot2)=\omega^2+\omega$. An ordinal$\alpha$is (additively) indecomposable iff$\alpha\gt 0$and whenever$\beta,\gamma\lt\alpha$, then$\beta+\gamma\lt \alpha$. One can easily check that the indecomposable$\alpha$are precisely those of the form$\omega^\beta$,$\beta\gt 0$. Say that$\alpha$is multiplicatively indecomposable iff$\alpha>0$and$\beta\gamma\lt \alpha$whenever$\beta,\gamma\lt\alpha$. Then$\alpha$is multiplicatively indecomposable iff it is$1$or has the form$\omega^{\omega^\beta}$. Ok, we can now state the remark; unfortunately I would not know who to credit for this observation, I think of it as folklore: An ordinal$\alpha$is multiplicatively indecomposable iff it is closed under Gödel pairing, that is,$\Gamma(\beta,\gamma)\lt\alpha$whenever$\beta,\gamma\lt\alpha$. In particular,$\Gamma(\kappa,\kappa)=\kappa$for any infinite cardinal$\kappa$, which of course implies that$\kappa\times\kappa$and$\kappa$have the same size. Also, if$\kappa$is uncountable, then there are$\kappa$ordinals$\alpha$below$\kappa$such that$\Gamma(\alpha,\alpha)=\alpha$. Of course, all of this works well in$\mathsf{ZF}$and all the definitions involved are absolute. I prefer a different approach when verifying that$\kappa\times\kappa$and$\kappa$have the same size, one that (again) is absolute and goes through in$\mathsf{ZF}$, but only requires the use of additively indecomposable ordinals: One first checks that there is a (recursive) bijection$h:\omega\times\omega\to\omega$with$h(0,0)=0$. Then, given ordinals$\alpha,\beta$, use their Cantor's normal form to write them as $$\alpha= \omega^{\alpha_1}n_1 + \omega^{\alpha_2}n_2 + \dots + \omega^{\alpha_k}n_k$$ and $$\beta= \omega^{\alpha_1}n'_1 + \omega^{\alpha_2}n'_2 + \dots + \omega^{\alpha_k}n'_k$$ where$\alpha_1 \gt \alpha_2 \gt \dots \gt \alpha_k$are ordinals, and$n_1,\dots,n_k, n'_1,\dots,n'_k$are natural numbers. (Note that these representations are not unique, but at least one of$n_i$and$n_i'$is non-zero iff$\alpha_i$appears as an exponent in the canonical form of$\alpha$or$\beta$). Now set $$H(\alpha,\beta)=\omega^{\alpha_1}h(n_1,n'_1)+\omega^{\alpha_2}h(n_2,n'_2)+\dots+ \omega^{\alpha_k}h(n_k,n'_k).$$ Then$H$is a bijection between$\alpha\times\alpha$and$\alpha$whenever$\alpha$is indecomposable. And an easy inductive argument, appealing to the explicit proof of Schröder-Bernstein, allows us to use$H$to argue that there is, provably in$\mathsf{ZF}$, a class function that assigns to each infinite ordinal$\alpha$a bijection between$\alpha\times\alpha$and$\alpha$. (Of course, the existence of this class function can also be argued from$\Gamma$, using that there are$\kappa$ordinals$\alpha$below$\kappa$with$\Gamma(\alpha,\alpha)=\alpha\$, but this second approach is somewhat easier.)