$i = 0
"$($i++)" #will output nothing
"$(($i++))" #will output the old value of $i
Additionally, typing $i++ and ($i++) directly on the console will do the same.
Why? I still don't exactly understand what $() does.
$(...), the subexpression operator is primarily useful in expandable (double-quoted) strings, "...", which is indeed what you're trying to use it for.
It does not modify the (often implicit) output behavior of the pipeline(s) it encloses.
++$i and $i++ are specialized assignment statements, which, like all assignment statements, do not produce pipeline output, but the assigned values can act as expression operands.
In order for regular, =-based assignments to participate in an expression, they need to be enclosed in (...), the grouping operator, which for ++$i and $i++ isn't strictly necessary; e.g., ++$i + 1
Using an (...)-enclosed assignment by itself in a pipeline also outputs the assignment value; that is, it passes the value being assigned through to the pipeline.
(...) or participating in an expression, pass the variable's current value through, before updating the variable with the incremented/decremented value.Therefore, increment / decrement assignments (++ / --) too must be enclosed in (...) in order to produce output: (++$i) and ($i++)
Inside "...", you must therefore use both (...) and $(...):
$i = 0
"$((++$i))" # -> 1
"$(($i++))" # -> 1, but $i contains 2 afterwards.
For a more comprehensive overview of PowerShell's output behavior, see this answer.