Question: Suppose that $f$ is differentiable on $I$. Use the product rule and mathematical induction to show that $f^n$ (the function f is raised to the nth power) is differentiable on $I$ for every positive integer $n$ and find a formula for $(f^n)’$.

Answer: 
I think other replies to this thread are confused about what the question is asking. The question is not asking to prove that $f^{(n)}$ is differentiable, it is asking to prove that $f^n$ (the $n^{th}$ power of $f$) is differentiable.

We'll use strong induction to show that $(f^n)' = nf^{n-1}f'$.

Consider the base case, $n=1$. $f^1 = f$ is differentiable by assumption of the problem, and the formula for its derivative is given by $f' = 1f^{1-1} f'$ because $f^{1-1} = 1$. So clearly our property holds.

Now we'll assume that $f^i$ is differentiable for any $i < n$ and its derivative moreover is given by $(f^i)' = if^{i-1}f'$ and we'll prove the same holds for $f^{n}$. We note that $f^{n} = f^{n-1} f$. From here, we can invoke the $i=2$ case (which is simply a statement of the product rule) to get that $f^n$ is differentiable. We can also use the $i=2, i=1$, and $i = n-1$ cases to get that $$(f^{n})' = [f^{n-1} f]' = (f^{n-1})'f + f' f^{n-1} = (n-1)f^{n-2}f'f + f'f^{n-1} = (n-1)f^{n-1}f' + f^{n-1}f' = nf^{n-1}f'$$ which is the formula we posited. So in particular, for every natural number $n$, we've proved by strong induction that $$(f^n)' = nf^{n-1}f'.$$
This is false as stated, if by $f^n$ you meant iterated derivative $f$ . Take $f(x)$ with: $$f(x)=-x^2/2$$ if $x\leq 0$ and $$f(x)=x^2/2$$ if $x>0$. Then $f$ is differentiable, but $f'=|x|$ is not.
( Note that if instead of a Real-variable function $f(x)$, you had $f(z)$ as a complex-valued function , then it would be true that $f^n(z)$ , as the $n-th$ derivative always exists--and it is analytic.

EDIT: I read $f^n$ to mean the $n-th$ application of the derivative, but some have poined out that is not what the OP meant. If by $f^n$ you mean n-th product, then you can say that when $f,g$ are differentiable, then $f.g$ is differentiable, and 
$(f.g)'=f'g+fg'$ , so that $f.f$ is differentiable. Then $(f^3)'$=$(f.f^2)'$ is also the product of differentiable functions, and the same applies. Then, using induction,
 $(f^{n+1})'=([(f^n)(f)]'$ is the product of differentiable functions.