I carry on following what is written at the 9th chapter of "Introduction to Set Theory" by Karel Hrbacek and Thomas Jech
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1.3 Theorem
Let $\lambda$ an infinite cardinal, let $k_{\alpha<\lambda}$ be nonzero cardinal numbers, and let $k=sup[k_{\alpha<\lambda}]$. Then
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \sum_{\alpha<\lambda}k_\alpha=k*\lambda$.
Proof. On the one hand, $k_\alpha\le k$ for each $\alpha<\lambda$, and so $\sum_{\alpha<\lambda}k_\alpha\le\sum_{\alpha<\lambda}k=k*\lambda$. On the other hand, we notice that $\lambda=\sum_{\alpha<\lambda}1\le\sum_{\alpha<\lambda}k_\alpha.$ We also have $k\le\sum_{\alpha<\lambda}k_\alpha$: the sum $\sum_{\alpha<\lambda}k_\alpha$ is an upper bound of the $k_\alpha$'s and $k$ is the least upper bound. Now since both $k$ and $\lambda$ are $\le\sum_{\alpha<\lambda}k_\alpha$, it follows that $k*\lambda$, which is the greater of the two, is also $\le\sum_{\alpha<\lambda}k_\alpha$. The conclusion of Theorem 1.3 is now a consequence of the Cantor-Bernstein Theorem.
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Well I don't understand why since both $k$ and $\lambda$ are $\le\sum_{\alpha<\lambda}k_\alpha$, it follows that $k*\lambda$, which is the greater of the two, is also $\le\sum_{\alpha<\lambda}k_\alpha$. Could someone explain to me this formally?
Anyway it seems to me that it could be possible to prove it in this way: since $\lambda$ is an infinite cardinal, $k$ must have this property and then it's $k*\lambda=k+\lambda \le\sum_{\alpha<\lambda}k_\alpha+\sum_{\alpha<\lambda}k_\alpha=\sum_{\alpha<\lambda}k_\alpha$; is it correct?
Exercise.
Show if a and b are infinite cardinals, then a×b = max{a,b}.
That will answer your question.
This his comment "which is the greater of the two."