Why are trailing line breaks stripped when used as an argument (for echo)?

Question

When I run this…

echo "          



fdsfsd
dsfsdf



"  

I really get the output of all these lines.

However, if I use this output as an argument for echo again…:

echo "$(echo "



fdsfsd
dsfsdf



")"

…I do only get the leading new lines, not the trailing ones:

(4 empty lines)  




fdsfsd
dsfsdf

So: Why does this happen? (Note I explicitly quoted the output of the $(), so I thought that would prevent such things.) And how can I prevent this?

(tested with zsh)

BTW, same result if I use printf instead of echo – although the former obviously always misses the last line break echo would usually add.