To prove : If $f^n$ has a unique fixed point $b$ then $f(b)=b$

If $f: \mathbb R \to \mathbb R$ be a function such that for some $n_o \in \mathbb N$ , the $n_o$th iterate of $f$ has a unique fixed point $b$ , then how to prove that $f(b)=b$ ? I cant think of anything , please help . Thanks .

Suppose that $f(b)=a$. Then $$f^{(n_0)}(a)=f^{(n_0)}(f(b))=f^{(n_0+1)}(b)=f(f^{(n_0)}(b))=f(b)=a\ .$$ Thus $a$ is a fixed point of $f^{(n_0)}$, and by assumption $a=b$.

Hint: What is

$$f^{n_0}(f(b))\,?$$

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