Perl eating line one on -n commandline option flag - perl

Question

I've started playing around with perl and I'm trying to figure out what is wrong with telling perl to use a loop if I also provide a loop?
It looks like perl is getting confused with the same open file descriptors but what I don't get is why does it eat the first line?
perl -ne 'while (<>) { print $_; }'
Of course in this simple example, I can simply perl -ne '{print $_}' to arrive at the same functional logic.
But what I want to know is what is going wrong with the double loop that the first line disappears if yet another while (<>) { } gets wrapped?
$ perl -ne '{print $_}' hello
hello
hello
world
world
^C
$ perl -ne 'while (<>) { print $_; }'
hello
world
world
^C
Update: According to the answers what seems to be happening is that Perl is waiting on the first loop for STDIN input. Upon receiving input on STDIN, the input is assigned to the internal buffer $_ and the logic proceeds to the second loop where again it waits for new STDIN input. Upon receiving new STDIN input, it clobbers the STDIN buffer $_ with the new STDIN input and begins printing.

Answer 1

You can itself check the code generated by one-liner using O=Deparse.
First:
$ perl -MO=Deparse -ne 'print $_;' file
LINE: while (defined($_ = <ARGV>)) {
print $_;
}
-e syntax OK
Second:
$ perl -MO=Deparse -ne 'while (<>) { print $_; }' file
LINE: while (defined($_ = <ARGV>)) {
while (defined($_ = <ARGV>)) {
print $_;
}
}
-e syntax OK
Now, It is easy to know what is wrong with second case. Outer while eats the first line of file and it is lost.

Answer 2

The -n flag wraps your code inside a while (<>) { ... } construct.
So in your second example, the code that is actually executed is
while (<>) # reads a line from STDIN, places it in $_
{
# you don't do anything with the contents of $_ here
while (<>) # reads a line from STDIN, places it in $_, overwriting the previous value
{
print $_; # prints the contents of $_
}
}
Which means the line that was read by the first <> is just lost.