bash - Why does ((count++)) return 1 exit code first time run

Question

I have no idea why the indicated line below is returning 1 while the subsequent executions of ((count++)) are returning 0.

[me@server ~]$ count=0
[me@server ~]$ echo $?
0
[me@server ~]$ count++
-bash: count++: command not found
[me@server ~]$ (count++)
-bash: count++: command not found
[me@server ~]$ ((count++))
[me@server ~]$ echo $?
1 <------THIS WHY IS IT 1 AND NOT 0??
[me@server ~]$ ((count++))
[me@server ~]$ echo $?
0
[me@server ~]$ ((count++))
[me@server ~]$ echo $?
0
[me@server ~]$ echo $count
3

Answer 1

See the excerpt from the help let page,

If the last ARG evaluates to 0, let returns 1; 0 is returned otherwise.

Since the operation is post-increment, ((count++)), for the very first time 0 is retained, hence returning 1

Notice, the same does not happen for pre-increment ((++count)), since the value is set to 1, on the first iteration itself.

$ unset count
$ count=0
$ echo $?
0
$ ++count
-bash: ++count: command not found
$ echo $?
127
$ ((++count))
$ echo $?
0