Clarification about Infinite Sums in Jech's Set Theory 3rd Edition On page 57 in the 3rd Edition of Jech's Set Theory, he begins with the elaboration on cardinal exponentiation. To that end the Hausdorff formula is introduced by noticing that for a regular cardinal $\kappa$,
\begin{equation}
  \kappa^\lambda = \cup_{\alpha < \kappa} \alpha^\lambda,
\end{equation}
with $\lambda$ denoting an infinite cardinal. From this Jech-apparently obviously-concludes that
\begin{equation}
  \kappa^\lambda = \sum_{\alpha < \kappa} |\alpha|^\lambda.
\end{equation}
My question is about the above conclusion.
On page 52 infinite sums are defined for disjoint families. This prerequisite appears not to be met by $\{\alpha^\lambda\}_{\alpha < \kappa}$, since $\beta < \gamma \implies \beta^\lambda \subseteq \gamma^\lambda$. My question is therefore why Jech applies the definition of inifinite sums, despite $\{\alpha^\lambda\}_{\alpha < \kappa}$ not being a disjoint family.
Appendix: Unless I have completely misunderstood something, I would guess that the answer to my question lies in the axiom of choice, i.e., we could replace $\{\alpha^\lambda\}_{\alpha < \kappa}$ with a disjoint family $\{X_\alpha\}_{\alpha < \kappa}$, $|X_\alpha| = |\alpha|$. Is this the right line of thinking?
 A: Using the regularity of $\kappa$ and the fact that the infinite cardinal $\lambda$ is less than $\kappa$, Jech notes that 
 $$ \kappa^\lambda=\bigcup_{\alpha<\kappa}\alpha^\lambda, $$
where both sides are understood as sets of functions. From this, he claims that, as cardinals, 
 $$ \kappa^\lambda=\sum_{\alpha<\kappa}|\alpha|^\lambda. $$
This would be obvious if the sets $\alpha^\lambda$ were disjoint as $\alpha$ varies but, as you point out, this is clearly not the case.
You suggest instead to replace these sets by disjoint copies, say replacing $\alpha$ by $\hat\alpha=\{\alpha\}\times\alpha$ for each $\alpha<\kappa$. This changes the union $\bigcup_\alpha\alpha^\lambda$ into the disjoint union $\bigcup_\alpha\hat\alpha^\lambda$. Note that there is an obvious injection from the former into the latter: Given any function $f\in\bigcup_\alpha\alpha^\lambda$, find the least $\alpha$ such that $f\!:\lambda\to\alpha$, and map $f$ to its copy in $\hat\alpha^\lambda$. 
This means that 
 $$ \kappa^\lambda\le\sum_{\alpha<\kappa}|\alpha|^\lambda. $$
However, Jech is claiming more, namely, equality rather than the inequality we just showed. 
Luckily for us, the inequality 
 $$ \sum_{\alpha<\kappa}|\alpha|^\lambda\le\kappa^\lambda $$
is quite easy to establish: First, for each $\alpha<\kappa$, $|\alpha|^\lambda\le\kappa^\lambda$, so $\sum_{\alpha<\kappa}|\alpha|^\lambda \le \sum_{\alpha<\kappa}\kappa^\lambda=\kappa\cdot\kappa^\lambda=\kappa^\lambda$. Note that this is the only place where we used that $\lambda$ is infinite (in fact, it suffices that $\lambda>0$).
From the two inequalities, the claimed equality now follows. Let me close by pointing out that this is a really useful heuristic: just as equality of sets is two containments, many equalities in cardinal arithmetic are really two inequalities. 
In the case at hand, one of the inequalities turned out to be obvious upon inspection. This is actually not so uncommon. The strategy you suggested, of replacing a union by a disjoint union so that its cardinality can be estimated via Jech's definition of infinite sums of cardinals, turns out to be very useful in practice.
A: The formula $\cup_{a<l}(a^l)$ is not a presentation of a sum of ordinals. It is the union of  a set of cardinal ordinals. (And therefore  it  $is$ a cardinal ordinal.)
The statement $k^l=\cup_{a<k}(a^l)$ (when  $k$ is an infinite regular cardinal ordinal) and $l$ is an infinite cardinal ordinal)) means (i) $\forall a<k\;(a^l\leq k^l)$ and  (ii) $\forall x\in k^l\;\exists a<k \;(x\in a^k).$
In cardinal arithmetic, when $a$ and $b$ are cardinals then $a^b$ is the cardinal of the set of functions from $b$ to $a$ . So (i) above is obvious.
