Use the product rule and induction (but NOT the chain rule) to prove that if $f(x)$ is a differentiable function, then for any $n \ge 1$, $$\frac{\mathrm d}{\mathrm dx} (f(x))^n = n(f(x))^{n−1} \times f'(x)$$
I have: base case $n=1$ so:
$$\begin{align} \frac{\mathrm d}{\mathrm dx} (f(x))^1 &= 1(f(x))^{1-1} \times f'(x) \\ \frac{\mathrm d}{\mathrm dx} f(x) &= f'(x) \end{align}$$
Do I then assume this works for the following? $$\frac{\mathrm d}{\mathrm dx} (f(x))^k = k(f(x))^{k−1} \times f'(x)$$
Then do I show…? $$\frac{\mathrm d}{\mathrm dx} (f(x))^{k+1} = (k+1)(f(x))^{k+1-1} \times f'(x)$$
I've tried doing that and "substituting" what $\frac{\mathrm d}{\mathrm dx} (f(x))^k$ is but it just makes a bigger mess and starts producing powers of $f'(x)$ everywhere. Then again maybe there's a trick to this that I have no idea about.
Using only the product rule, here is the induction step (for $n\ge2$).
Set $g(x)=f(x)^{n-1}$; by the induction hypothesis, $g'(x)=(n-1)f'(x)f(x)^{n-2}$ and \begin{align} D(f(x)^n) &=D(f(x)g(x))\\[4px] &=f'(x)g(x)+f(x)g'(x)\\[4px] &=f'(x)f(x)^{n-1}+f(x)\cdot (n-1)f'(x)f(x)^{n-2}\\[4px] &=f'(x)f(x)^{n-1}+(n-1)f'(x)f(x)^{n-1}\\[4px] &=nf'(x)f(x)^{n-1} \end{align}
You can supply the base case for $n=2$.