Running an awk command with $SHELL -c returns different results

I am trying to use awk to print the unique lines returned by a command. For simplicity, assume the command is ls -alh.

If I run the following command in my Z shell, awk shows all lines printed by ls -alh

ls -alh | awk '!seen[$0]++'

However, if I run the same command with $SHELL -c while escaping the ! with backslash, I only see the first line of the output printed.

$SHELL -c "ls -alh | awk '\!seen[$0]++'"

How can I ensure the latter command prints the exact same outputs as the former?

EDIT 1:

I initially thought the ! could be the issue. But changing the expression '!seen[$0]++' to 'seen[$0]++==0' has the same problem.

EDIT 2:

It looks like I should have escaped $ too. Since I do not know the reason behind it, I will not post an answer.

In the second form, $0 is being treated as a shell variable in the double-quoted string. The substitution creates an interestingly mangled awk command:

> print $SHELL -c "ls -alh | awk '\!seen[$0]++'"
/bin/zsh -c ls -alh | awk '!seen[-zsh]++'

The variable is not substituted in the first form since it is inside single quotes.

This answer discusses how single- and double-quoted strings are treated in bash and zsh:
https://stackoverflow.com/questions/6697753/difference-between-single-and-double-quotes-in-bash

Escaping the $ so that $0 is passed to awk should work, but note that quoting in commands that are parsed multiple times can get really tricky.

> print $SHELL -c "ls -alh | awk '\!seen[\$0]++'"
/bin/zsh -c ls -alh | awk '!seen[$0]++'