The Lambda Calculus Y Combinator (Requesting Confirmation or Clarification)

So basically it is f(f(f(do stuff here or put value here or both))) with more parenthesis as it runs or is it f(((do stuff here or put values here or both))) with more parenthesis as it runs? Basically looks like this f(((((((λx. f (x x)) (λx. f (x x)))))))) with growing numbers of parenthesis or are there f's between the left parenthesis? Currently believe it is likely f(f(f(f and so on...AKA: "function with same function as input here: (function with same function as input here: (function with same function as input here:" and so on. Thanks in advance for confirming and clarifying. PS: Can confirm how the y combinator gets past step 3 in the following list of iterations? IE: Does the input function get passed in as f again somehow or is the f on left side of the function always automatically assigned the original input value on every pass?: Step 1: (λf. (λx. f (x x))(λx. f (x x)))SomeFunction Step 2: (λx. SomeFunction (x x))(λx. f (x x)) Step 3: SomeFunction(λx. f (x x))(λx. f (x x)) Step 4 (Want to make sure this "f is automatically assigned SomeFunction as val on every pass" step is correct): SomeFunction(SomeFunction(λx. f (x x))(λx. f (x x)))