How do I write an arrow function in ES6 recursively?

Question

Arrow functions in ES6 do not have an arguments property and therefore arguments.callee will not work and would anyway not work in strict mode even

Answer 1

I found the provided solutions really complicated, and honestly couldn't understand any of them, so i thought out a simpler solution myself (I'm sure it's already known, but here goes my thinking process):

So you're making a factorial function

x => x < 2 ? x : x * (???)

the (???) is where the function is supposed to call itself, but since you can't name it, the obvious solution is to pass it as an argument to itself

f => x => x < 2 ? x : x * f(x-1)

This won't work though. because when we call f(x-1) we're calling this function itself, and we just defined it's arguments as 1) f: the function itself, again and 2) x the value. Well we do have the function itself, f remember? so just pass it first:

f => x => x < 2 ? x : x * f(f)(x-1)
                            ^ the new bit

And that's it. We just made a function that takes itself as the first argument, producing the Factorial function! Just literally pass it to itself:

(f => x => x < 2 ? x : x * f(f)(x-1))(f => x => x < 2 ? x : x * f(f)(x-1))(5)
>120

Instead of writing it twice, you can make another function that passes it's argument to itself:

y => y(y)

and pass your factorial making function to it:

(y => y(y))(f => x => x < 2 ? x : x * f(f)(x-1))(5)
>120

Boom. Here's a little formula:

(y => y(y))(f => x => endCondition(x) ? default(x) : operation(x)(f(f)(nextStep(x))))

For a basic function that adds numbers from 0 to x, endCondition is when you need to stop recurring, so x => x == 0. default is the last value you give once endCondition is met, so x => x. operation is simply the operation you're doing on every recursion, like multiplying in Factorial or adding in Fibonacci: x1 => x2 => x1 + x2. and lastly nextStep is the next value to pass to the function, which is usually the current value minus one: x => x - 1. Apply:

(y => y(y))(f => x => x == 0 ? x : x + f(f)(x - 1))(5)
>15