The version I was thinking when I wrote the comment is simpler
type Flatten<T> = T extends Array<infer U> ? Flatten<U> : T
The average X dev in Y language doesn't need to understand Z is a poor argument in the context of writing better software.
as a person that never touched JS and TS... what's the difference between the two answers?
For one, the simple answer is incomplete. It gives the fully unwrapped type of the array but you still need something like
type FlatArray<T extends unknown[]> = Flatten<T[number]>[]
First one flattens a potentially-nested tuple type. E.g., FlatArr<[number, [boolean, string]]> is [number, boolean, string].
Second one gets the element type of a potentially-nested array type. E.g., Flatten<number[][]> is number.
For what it's worth, I've never needed to use either of these, though I've occasionally had other uses for slightly fancy TypeScript type magic.
> The average X dev in Y language doesn't need to understand Z is a poor argument in the context of writing better software.
It's a good response to the claim that we'd be surprised at how many would fail to do this, though.