If $\alpha$ is an indecomposable ordinal, why $\Gamma(\alpha\times{\alpha})=\alpha$? This is from exercise 3.4 of Thomas Jech's "Set Theory", stated:

"Show that $\Gamma(\alpha\times{\alpha})\leq{\omega^{\alpha}}$. Thus $\Gamma(\alpha\times{\alpha})=\alpha$ for all indecomposable ordinals $\alpha$".

$\Gamma$ is the Gödel pairing function on ordinals. An ordinal $\alpha$ is indecomposable if there are no $\beta<\alpha$ and $\gamma<\alpha$ such that $\beta+\gamma=\alpha$.
The first part is trivial, since it holds for $\alpha=0$, and the function $\omega^{\alpha}$ is normal in the variable $\alpha$.
I'm having trouble to show the second part, the only connection I see is the fact that $\gamma$ is indecomposable if and only if $\gamma=\omega^{\alpha}$ for some $\alpha$, which is easily proven from Cantor's normal form theorem.
Any help will be appreciated.
 A: This is also an exercise in Kunen 2011 (p. 67) which I am studying at the moment. Here is how I proved it. Kunen's version of the exercise is as follows:
Exercise I.11.7. For $\alpha \geq \omega$, show that $\mbox{type}(\alpha \times \alpha, \lhd) = \alpha$ iff  $\alpha = \omega^{\mu}$, where $\mu$ is 
either $1$ or an infinite decomposable ordinal. 
Here $\lhd$ is the canonical well ordering on ONxON. Note that Kunen introduces decomposable ordinals in a previous exercise:
Exercise I.9.53. Let $\gamma$ be a limit ordinal. Show that the following are equivalent:


*

*$\forall \alpha,\beta < \gamma \:\: [\alpha + \beta < \gamma]$.

*$\forall \alpha < \gamma \:\: [\alpha + \gamma = \gamma]$.

*$\forall X \subseteq \gamma \:\: [\mbox{type}(X)=\gamma \vee \mbox{type}(\gamma\setminus X)=\gamma]$.

*$\exists \delta \:\: [\gamma = \omega^\delta]$.  $\blacksquare$
Such $\gamma$ are called (additively) indecomposable.
Before proving Exercise I.11.7, I proved the following proposition, the proof of which I will not reproduce here.
Proposition. For any ordinal $\alpha> 2$, the following conditions are equivalent:


*

*$\beta\cdot\alpha = \alpha$ for all non-zero $\beta<\alpha$.

*$\xi\cdot\eta < \alpha$ for all $\xi, \eta < \alpha$.

*$\alpha$ is of the form $\omega^{\omega^{\delta}}$ for some ordinal $\delta$. $\blacksquare$
An ordinal $\alpha$ satisfying the above conditions is called multiplicatively indecomposable. We can use this terminology to restate the exercise as follows:
Exercise I.11.7. For $\alpha \geq \omega$, show that $\mbox{type}(\alpha \times \alpha, \lhd) = \alpha$ iff $\alpha$ is multiplicatively indecomposable.
Warm-up
Before proving the exercise, I tried to get better acquainted with the function
$$\Gamma (\alpha , \beta) \: = \: \mbox{type}\{(\xi,\eta) : (\xi,\eta) \lhd (\alpha , \beta) \}.$$
Observe that 
$$\Gamma (0,\alpha) = \mbox{type}(\alpha \times \alpha, \lhd).$$
So the exercise I.11.7. directly involves the function $\Gamma$.
For example, for finite ordinals $n\in \omega$ we have
$$\Gamma (0,n)  = \mbox{type}(n \times n, \lhd) = n^2. $$
The function $\Gamma$ behaves at successor steps in the second coordinate as follows:
$$\Gamma (\alpha , \beta+1) = \left\{   
\begin{array}{ll}
  \Gamma (\alpha , \beta) + 1  & \mbox{if } \alpha > \beta  \\
  \Gamma (\alpha , \beta) + \beta  & \mbox{if } \alpha = \beta  \\
  \Gamma (\alpha , \beta) + \gamma + \beta + 1 + \alpha  & \mbox{if } \alpha < \beta  
\end{array} \right.   $$
where $\gamma$ in the third case is the unique ordinal satisfying $\alpha + \gamma = \beta$.
One can prove this formula by writing down the chain of $\lhd$-successors from $(\alpha, \beta)$ to $(\alpha, \beta+1)$. The first case is obvious since 
if $\alpha > \beta$ then $(\alpha, \beta+1)$ is the $\lhd$-successor of $(\alpha, \beta)$. The second case follows from the chain of $\lhd$-successors
$$ (\beta,\beta) \: \lhd \:  (0,\beta+1) \: \lhd \: (1,\beta+1) \: \lhd \:  (2,\beta+1) \: \lhd \: \cdots \: \lhd \: (\beta,\beta+1).$$
In the third case, the chain is as follows:
\begin{align*}
  (\alpha, \beta) & \: \lhd \:  (\alpha+1, \beta) \: \lhd \: (\alpha+2, \beta)  \: \lhd \: \cdots \: \lhd \: (\xi,\beta) \: \lhd \: \cdots   \\
    \mbox{} & \: \lhd \: (\beta, 0)  \: \lhd \: (\beta, 1) \: \lhd \: \cdots \: \lhd  \: (\beta, \beta)             \\   
    \mbox{} & \: \lhd \: (0,\beta +1)  \: \lhd \: (1,\beta +1) \: \lhd \: \cdots \: \lhd \: (\alpha, \beta+1) 
\end{align*}
The $\xi$ in the first line takes values $\alpha < \xi <\beta$.
By considering the $\lhd$-predecessors when $\lambda$ is limit, one can get
$$\Gamma (\alpha , \lambda) = \left\{   
\begin{array}{ll}
  \sup_{\xi < \lambda} \Gamma (\alpha , \xi)  & \mbox{if } \alpha \geq \lambda  \\
  \sup_{\xi < \lambda} \Gamma (\alpha , \xi) \: + \: \alpha & \mbox{if } \alpha < \lambda  
\end{array} \right.   $$
We will make use of the following particular instances of the above formulas:
\begin{align*}
  \Gamma(0, \beta +1) & =  \Gamma(0,\beta) + \beta + \beta + 1  \\
 & \\
  \Gamma(0, \lambda ) & =  \sup_{\xi < \lambda} \Gamma (0 , \xi)  \:\:\:\:\: \mbox{ if } \lambda \mbox{ is limit}
\end{align*}
Kunen also remarks that (p. 67)
$$\Gamma (0,\alpha) = \mbox{type}(\alpha \times \alpha, \lhd) \geq \alpha .$$
If $(\alpha,\beta)+1$  is the $\lhd$-successor of $(\alpha,\beta)$ then
$$ (\alpha,\beta)+1 \: = \: \left\{   
\begin{array}{ll}
  (0,\beta+1) & \mbox{ if } \alpha = \beta  \\
  (\alpha+1,\beta) & \mbox{ if } \alpha < \beta  \\
  (\alpha,\beta+1) & \mbox{ if } \alpha > \beta. 
\end{array} \right.   $$
Let us call $(\delta,\gamma)$ limit if
$$\forall (\xi,\eta)\in\text{ON}\times\text{ON}  \quad 
\big[(\xi,\eta)\lhd (\delta,\gamma) \rightarrow  (\xi,\eta)+1 \lhd (\delta,\gamma) \big].$$
Then 
$$(\delta, \gamma) \text{ is limit } \quad \text{ iff } \quad \gamma\text{ is limit or } [\delta < \beta \wedge \delta \text{ is limit}].$$
Proof of the Exercise
Lemma. For all ordinals $\alpha$ we have $\Gamma(0,\alpha) < \alpha \cdot \alpha + 1$.
Proof. This lemma holds clearly for finite $\alpha$: If $n\in \omega$, then
$$\Gamma (0,n) = \mbox{type}(n \times n, \lhd) = n^2 < n\cdot n +1.$$
Let us prove the lemma by induction on $\alpha$. 
We may assume that $\alpha \geq \omega$.
If $\alpha = \beta +1$ for some $\beta$ and $\Gamma(0,\beta)< \beta\cdot\beta + 1$ then
\begin{align*}
 \Gamma(0,\beta + 1) & \: = \: \Gamma(0,\beta) + \beta + \beta + 1 \: \leq \: \beta\cdot \beta + 1 + \beta + \beta + 1 \: = \: \beta\cdot \beta + \beta + \beta + 1 \\
                     & \: = \: (\beta +1)\cdot (\beta +1) \: < \: (\beta +1)\cdot (\beta +1)+1
\end{align*}
If $\alpha$ is limit and $\Gamma(0,\xi)<\xi\cdot \xi +1$ for all $\xi <\alpha$ then
$$\Gamma(0,\alpha) \: =  \: \sup_{\xi < \lambda} \Gamma(0,\xi) \: \leq \:  \sup_{\xi < \lambda} (\xi\cdot \xi +1) \: \leq \: \alpha\cdot \alpha \: < \: \alpha\cdot \alpha +1.$$
This completes the induction. $\blacksquare$
Lemma. If $\alpha < \omega^{\delta}$ then $\Gamma(0,\alpha) < \omega^{\omega^{\delta}}$.
Proof. Assume that $\alpha < \omega^{\delta}$. Then $\alpha < \omega^{\omega^{\delta}}$ as well. 
Since $\omega^{\omega^{\delta}}$ is multiplicatively indecomposable, 
$\alpha\cdot \alpha < \omega^{\omega^{\delta}}$. As $\omega^{\omega^{\delta}}$ is limit 
we have $\alpha\cdot \alpha + 1 < \omega^{\omega^{\delta}}$. Thus
$\Gamma(0,\alpha) < \alpha\cdot \alpha + 1 < \omega^{\omega^{\delta}}$. $\blacksquare$
This lemma gives us one direction of the exercise:
Lemma. $\Gamma (0, \omega^{\omega^{\delta}}) = \omega^{\omega^{\delta}}.$
Proof.  $\Gamma (0, \omega^{\omega^{\delta}}) \: = \:  \sup_{\xi < \omega^{\omega^{\delta}}} \Gamma (0, \xi) \:  \leq \: 
\sup_{\xi < \omega^{\omega^{\delta}}} \omega^{\omega^{\delta}} \:  = \:  \omega^{\omega^{\delta}}$. $\blacksquare$
The following inequality will establish the other direction of the exercise.
Lemma. $\alpha\cdot \alpha < \Gamma(0,\alpha + \alpha) + 1$.
Proof.  We will show the lemma by induction on $\alpha$. Since $n\cdot n < (n+n)^2 + 1$ for $n\in\omega$, 
this lemma is true for finite ordinals. From now on we may assume that $\alpha \geq \omega$.
Assume that $\alpha=\beta +1 $ for some $\beta$ and $\beta\cdot\beta < \Gamma(0,\beta+\beta) +1$. Then
\begin{align*} \Gamma(0,(\beta+1)+(\beta+1)) & \: = \: \Gamma(0,\beta+\beta+1) \: = \: \Gamma(0,\beta+\beta) + \beta + \beta +\beta + \beta + 1  \\
   & \geq \: \beta\cdot \beta + \beta + \beta+ \beta+ \beta +  1 \: \geq \: (\beta+1)\cdot (\beta+1).
\end{align*}
So $ (\beta+1)\cdot (\beta+1) < \Gamma(0,(\beta+1)+(\beta+1)) +1 $.
Now assume that $\alpha$ is limit and $\xi\cdot\xi < \Gamma(0,\xi+\xi)+1$ holds for all $\xi<\alpha$. Then
$$ \Gamma(0,\alpha+\alpha) \: = \: \sup_{\xi<\alpha+\alpha} \Gamma(0,\xi) \: \geq \:  \sup_{\alpha<\xi<\alpha+\alpha} \Gamma(0,\xi) \: \geq \: \sup_{\alpha<\xi<\alpha+\alpha} \xi 
 \: = \: \alpha+\alpha.$$
Hence $\alpha + \alpha < \Gamma(0,\alpha + \alpha) + 1$. This completes the induction. $\blacksquare$
Finally, we have:
Lemma. If $\Gamma(0,\alpha) = \alpha$ then $\alpha$ is multiplicatively indecomposable.
Proof. Assume that $\Gamma(0,\alpha) = \alpha$ . Clearly, $\alpha$ is a limit ordinal. 
It suffices to show that if $\gamma < \alpha$ then $\gamma\cdot\gamma < \alpha$. Since
$$\gamma + \gamma \: < \: \Gamma(0,\gamma) + \gamma + \gamma +1 \: = \: \Gamma(0,\gamma+1) \: < \: \Gamma(0,\alpha) \: = \: \alpha$$
we have $\gamma + \gamma < \alpha$ and 
$$ \Gamma(0,\gamma+\gamma) \: < \: \Gamma(0,\alpha) \: = \: \alpha.$$
As $\alpha$ is a limit, we have $\Gamma(0,\gamma+\gamma) + 1 < \alpha$ as well. Thus
$$ \gamma+\gamma \: < \: \Gamma(0,\gamma+\gamma) +1  \: < \: \alpha$$
and $\alpha$ is multiplicatively indecomposable. $\blacksquare$
