Question 94
Suppose function $f: \mathbb{N} \to \mathbb{N}$ satisfies
$$
f^{bf(a)}(a+1) = (a+1)f(b)
$$
for all positive integers $a, b \in \mathbb{N}$, where $f^k(n) = f(f(\cdots f(n)\cdots))$ denotes the composition of $f$ with itself $k$ times. Prove that $f(n) = n+1$ for all $n \in \mathbb{N}$.
$$
f^{bf(a)}(a+1) = (a+1)f(b)
$$
for all positive integers $a, b \in \mathbb{N}$, where $f^k(n) = f(f(\cdots f(n)\cdots))$ denotes the composition of $f$ with itself $k$ times. Prove that $f(n) = n+1$ for all $n \in \mathbb{N}$.