In what sense $\alpha \times \alpha$ is the initial segment generated by $(0,\alpha)$ in $Ord \times Ord$? This is from Jech's book on set theory:

We define a well ordering of the class $Ord \times Ord$ of ordinal pairs.
  Under this well ordering, each $\alpha \times \alpha$ is an initial segment of $Ord^2$. Moreover, the well ordered class $Ord^2$ is isomorphic to the class $Ord$ and we have a one-to-one function $\Gamma$ of $Ord^2$ onto $Ord$:
  We define
\begin{align}(\alpha,\beta) < (\gamma,\delta)  \iff&  
\max\{ \alpha, \beta \} < \max\{\gamma,\delta\},\\ &\text{or }
\max\{ \alpha, \beta \} < \max\{\gamma,\delta\}\text{ and }\alpha < \beta,\\& \text{or }\max\{ \alpha, \beta \} = \max\{\gamma,\delta\}\text{ and }\alpha = \beta\text{ and }\beta < \delta.\end{align}
The relation $<$ defined above, is a linear ordering of the class $Ord \times Ord$. Morover, if $X \subset Ord \times Ord$ is nonempty, then $X$ has a least element. Also, for each $\alpha$, $\alpha \times \alpha$ is the initial segment given by $(0,\alpha)$. 

I am trying to understand, how come $Ord \times Ord$ is isomorphic to $Ord$. How come the relation is one-to-one? For example:
Applying the above relation, If $\alpha \times \alpha$ is the initial segment given by $(0,\alpha)$,what will the initial segment given by $(\alpha,\alpha)$ represent in $Ord^2$?

It means that $\min(\mathsf{Ord\times Ord}\setminus(\alpha\times\alpha))=(0,\alpha)$. 
Note that all the pairs in $\alpha\times\alpha$ have both coordinates less than $\alpha$, so indeed $(0,\alpha)$ is above all of them in this order, and any pair $(\beta,\gamma)$ which is less than $(0,\alpha)$ is necessarily a pair such that $\beta$ and $\gamma$ are less than $\alpha$, and therefore $(\beta,\gamma)\in\alpha\times\alpha$.
