Solve a power tower of function compositions Lets begin with a simple functional equation ;)

Find all functions $f: \mathbb{N}\to\mathbb{N}$ st. $f(x)=x+1$ for all $x \in \mathbb{N}$.

I know what you are thinking, "math.SE is for professionals asking and working in their respective fields!"
Well, I was pondering the functional equation

Find all functions $f: \mathbb{N}\to\mathbb{N}$ st. $f^{f(y)}{x}=x+y+1$ for all $x,y \in \mathbb{N}$.

(where $f^a(x)$ means $f$ iterated $a$ times on $x$) I wanted to extend this equation:

Find all functions $f: \mathbb{N}\to\mathbb{N}$ st. $f^{f^{f(z)}(y)}(x)=x+y+z+1$ for all $x,y,z \in \mathbb{N}$.


Generalize: Find all functions $f: \mathbb{N}\to\mathbb{N}$ st. $f^{f\cdots {f^{f^{f^{f{(a_1)}}(a_2)}(a_3)}(a_4)}\cdots (a_{n-1})}(a_n)=a_1+a_2+\cdots+a_{n}+1$ $\forall a_1,a_2,\ldots a_n\in \mathbb{N}$.

Can anyone provide a full solution to the third and fourth problems proposed in this post? (Would prefer first a solution to problem 3, as I have been thinking about this for a month already and am excited to see if my answer is correct :) ) Thankz!
 A: I'll assume $\mathbb{N}$ refers to the positive integers. We'll solve the generalisation.
Let $f\cdots {f^{f^{f^{f{(a_1)}}(a_2)}(a_3)}(a_4)}\cdots (a_{n-1})=A$, $X=a_1+\dots + a_{n-1}$ and $a_n=a$ so that $$f^A(a)=a+X + 1 \tag{1}$$ for all such $A,X,a$.
Taking $f$ both sides of $(1)$ gives $f^{A+1}(a_n)=f(f^A(a))=f(a+X+1)$ and replacing $a$ with $f(a)$ in the same expression gives $f^{A+1}(a)=f^A(f(a))=f(a)+X+1$ hence $$f(a+X+1)=f(a)+X+1.$$ Note $X$ can take any integer value at least $n-1$, so taking $X=n+1$ gives $$f(a+n+2)=f(a)+n+2. \tag{2}$$ However, shifting $a\to a+1$ and taking $X=n$ gives $$f(a+n+2)=f(a+1)+n+1\tag{3}$$ whereupon comparing $(2)$ and $(3)$ gives $f(a)+1=f(a+1)$ for all $a\in \mathbb{Z}^+$. So $f(a_i)=a_i+c$ for fixed $c$.
Now with this definition of $f$, since $f^\ell(k)=k+c\ell$, we can explicitly compute $$A=a_{n-1}+ca_{n-2}+c^2a_{n-3} + \dots + c^{n-2}a_1+c^{n-1}$$ so the problem statement translates to $$a_n+ca_{n-1}+c^2a_{n-2}+\dots + c^{n-1}a_1+c^n=a_1+a_2+\dots + a_n+1\tag{4}$$ implying $c=1$. Hence the only satisfying function is $f(x)=x+1$ (which works as $(4)$ then is an equality).
