nonexistence of a function satisfying $f(x)^{f(y)}=y^x$

Question

prove that there is no function $f:\mathbb{N}\mapsto \mathbb{N}$ such that $\forall x, y \in \mathbb{N}\ (f(x))^{f(y)}=y^x$
My attempt: If such function exists, so for $x,y,z \in \mathbb{N}$ $(f(x))^{f(y+z)}=(y+z)^x$ but in the other hand $(f(x+y))^{f(z)}=z^{(x+y)}=z^xz^y=(f(x))^{f(z)}(f(y))^{f(z)}$
and I need a Hint to accomplish this.

Answer 1

Denote $a=f(1)$ then $a^a = 1^1 = 1$. Therefore $a=f(1)=1$ and $1 = 1^{f(2)} = 2^1=2$, a contradiction.