Get PID of command in one line bash script

When trying to get the PID of a shell command, it is possible using a script file:

#!/bin/bash

ls -lha &
echo "$!"

However, it is not possible using only the terminal. /bin/bash -c 'ls -lha & ; echo "$!"' This returns the error: > /bin/bash: -c: line 1: syntax error near unexpected token `;' The manual states the following: > A *list* is a sequence of one or more pipelines separated by one of the operators ;, &, &&, or ||, and optionally terminated by one of ;, &, or ``.  Of these list operators, && and || have equal precedence, followed by ; and &, which have equal precedence. > >A sequence of one or more newlines may appear in a *list* instead of a semicolon to delimit commands. > > If a command is terminated by the control operator &, the shell executes the command in the background in a subshell.  The shell does not wait for the command to finish, and the return status is 0.  Commands separated by a ; are executed sequentially; the shell waits for each command to terminate in turn.  The return status is the exit status of the last command executed. So I also tried using newline, but was not able to use it: /bin/bash -c 'ls -lha & \n echo "$!"' ### How can I call a shell command to run on background and get the PID? P.S.: I will not use ls command and it is a more complex script.  This is just a minimal example to reproduce the problem.