$f^n = f^k \circ f^{n-k}$ $f : A \to A$ and $n \in n$, Let $f^n$ be defined by $f^1 = f$ and $$f^n = f \circ f^{n-1}$$ for $n \gt 1$.
Let $n$ and $k$ be natural numbers with $k \lt n$. Prove $$f^n = f^k \circ f^{n-k}$$
Induction: $n=2$
$f^2 = f \circ f^{2-1} \implies f^2 = f^1 \circ f^1$
Hence, the base case holds true.
I.H: Suppose its true for $n=m$. We have to prove that it also holds true for $n=m+1$,
$$f^{m+1} = f \circ f^m$$
How do I show that it will be equal to $$f^n = f^k \circ f^{n-k}$$

Hint: fix $m$ and prove by induction on $n$ that $f^{n+m} = f^n \circ f^m$. 
