$\kappa\cdot\sum_{i\in I}\lambda_i=_c\sum_{i\in I} \kappa\cdot \lambda_i$ Moschovakis Exercise x4.20

Prove that for all indexed families of cardinals, $$\kappa\cdot\sum_{i\in I}\lambda_i=_c\sum_{i\in I} \kappa\cdot \lambda_i$$

We have $$\kappa\cdot\sum_{i\in I}\lambda_i=_c\kappa\times\{(i,x):i\in I,x\in\lambda_i\}\\
\sum_{i\in I} \kappa\cdot \lambda_i=_c\{(i,y):i\in I, y\in \kappa\cdot \lambda_i\}$$
and we need to establish a bijective correspondence between these sets. An element $(k,(i,x))$ of the first set must be mapped to an element $(i,y)$ of the second set. But how to construct $y\in \kappa\cdot\lambda_i=|\kappa\times \lambda_i|$ from $k$ and $x$? I can construct the element $(k,x)\in\kappa\times \lambda_i$, but $\kappa\times \lambda_i$ is not the same as $|\kappa\times \lambda_i|$ (see definitions in this question).
 A: By definition of cardinal multiplication and the definition of the cardinality operator $||$, there are bijections $\rho_i : \kappa \times\lambda_i\to \kappa\cdot \lambda_i $ and a bijection $\rho : \kappa \times \sum_i \lambda_i \to \kappa \cdot \sum_i\lambda_i.$ 
We have $$ \kappa\times \sum_i\lambda_i = \kappa\times \{(i,x):i\in I, x\in\lambda_i\}=\{(\alpha,(i,x)): \alpha\in\kappa, i\in I,  x\in\lambda_i\}$$ and $$ \sum_i\kappa\times \lambda_i = \{(i,z):i\in I, z\in\kappa\times\lambda_i\} = \{(i,(\alpha,x)):i\in I, \alpha\in\kappa, x\in\lambda_i\}.$$
It seems like it was already clear to you that there is a bijection $F$ between these two sets.
Finally, we have $$ \sum_i \kappa\cdot \lambda_i = \{(i,y):i\in I, y\in \kappa\cdot \lambda_i\}$$ and the function $\rho^*:\sum_i\kappa\times\lambda_i\to\sum_i\kappa\cdot\lambda_i$ that takes $ (i,z)\mapsto (i,\rho_i(z))$ is a bijection. 
Composing $\rho,$ $F$ and $\rho^*$ in the proper order (perhaps some of them need to be inverted) gives the bijection you want. 
